- For Section 20.3 do these problems 1, 3, 5, 7, 8, 11, 13, 14, 15-18, 20, 22, 23, 24, 25, 26, 27, 29. For Exercises 1 and 3 find a potential for the given vector field. For Exercises 7 and 11 find a vector potential for the given vector field. For Problem 25 find a potential for the vector field $\boldsymbol{E}.$
-
Let me point out few useful facts about $\operatorname{curl}.$
- For a differentiable scalar function $f$ we have that \[ \operatorname{curl}\bigl(\operatorname{grad} f\bigr) = \boldsymbol{0}. \] Therefore adding a gradient field to a vector field $\boldsymbol{F}$ will not change its curl: \[ \operatorname{curl} \boldsymbol{F} = \operatorname{curl}\bigl(\boldsymbol{F} + \operatorname{grad} f\bigr). \] Written with the coordinate functions this equality reads \[ \operatorname{curl} \langle F_1, F_2, F_3 \rangle = \operatorname{curl} \langle F_1 + f_x, F_2 +f_y , F_3 + f_z \rangle. \] In the above equality $f$ can be any scalar function.
- The fact from the previous item can be used to solve a problem like this: Find a vector field \[ \boldsymbol{G} = \langle G_1, G_2, 0 \rangle \] such that \[ \operatorname{curl} \boldsymbol{G} = \operatorname{curl}\bigl\langle 3 y z, 2 x z + y, x y + 2 z \bigr\rangle. \] Using the property from the previous item we need to find a scalar function $f$ such that \[ x y + 2 z + f_z = 0. \] There are many such functions $f.$ The simplest one is $f(x,y,z) = -xyz-z^2.$ Thus, one function $\boldsymbol{G}$ that satisfies the given requirements is \[ \boldsymbol{G} = \bigl\langle 2 y z, x z + y, 0 \bigr\rangle. \]
- Using the property from the first item we can develop a strategy for answering questions like this: Find a vector field $\boldsymbol{G}$ such that \[ \operatorname{curl} \boldsymbol{G} = \bigl\langle 2 y z, 5 y - x z, x - 5 z \bigr\rangle. \] By the first item we can look for $\boldsymbol{G}$ such that \[ \boldsymbol{G} = \langle G_1, G_2, 0 \rangle. \] Then we have the equality \[ \Bigl\langle -\frac{\partial G_2}{\partial z}, \frac{\partial G_1}{\partial z}, \frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} \Bigr\rangle = \bigl\langle 2 y z, 5 y - x z, x - 5 z \bigr\rangle. \] That is, we need to find two functions $G_1$ and $G_2$ which satisfy the partial differential equations: \[ -\frac{\partial G_2}{\partial z} = 2 y z, \quad \frac{\partial G_1}{\partial z} = 5 y - x z, \quad \frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} =x - 5 z. \] From the first and the second equations we get \[ G_2(x,y,z) = - yz^2 + a(x,y), \quad G_1(x,y,z) = 5yz -\frac{1}{2} xz^2 + b(x,y). \] Substituting these in the third differential equation we get \[ \frac{\partial a}{\partial x} - 5z - \frac{\partial b}{\partial y} = x - 5 z. \] There are many functions $a(x,y)$ and $b(x,y)$ which satisfy this equation. We just need one $a$ and one $b$. We can take \[ a(x,y) = \frac{1}{2} x^2, \quad b(x,y) = 0. \] Thus, we found out that \[ \boldsymbol{G} = \Bigl\langle 5yz -\frac{1}{2} xz^2, - yz^2 + \frac{1}{2} x^2, 0 \Bigr\rangle \] is a vector potential of $\bigl\langle 2 y z, 5 y - x z, x - 5 z \bigr\rangle.$ This can be easily verified by calculating the curl of $\boldsymbol{G}.$
- For Section 20.2 do these problems 1, 5, 7, 8, 9, 11, 13, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37. Notice that 31 is an interesting problem since both sides of Stoke's Theorem seems difficult to evaluate. However, it is much easier to evaluate the circulation. You only need to notice that all the terms in the Riemann integral obtained from the circulation integral are periodic with period $2\pi$ and either odd or even.
- We started Section 20.1 today. Do these problems 1, 3, 4, 5, 8, 9, 13, 14-17, 19, 21, 25, 27, 28, 29, 30, 31, 32, 33.
-
In this item I will discuss the following remarkable theorem:
Below is a polyhedron with six faces with the outward-pointing area vectors emphasized in different colors. The theorem claims that the sum of these area vectors is the zero vector.
The next animation provides a visual proof of the theorem for a tetrahedron.Place the cursor over the image to start the animation.
We use the picture below to give an algebraic proof of the theorem for a tetrahedron.
Next, we introduce notation for the colored vectors in the picture above. Denote the purple vector by $\vec{a},$ the olive vector by $\vec{b}$ and the orange vector by $\vec{c}.$ Then the pink vector is $\vec{a} - \vec{c}$ and the chartreuse vector is $\vec{b} - \vec{c}.$ Based on the definition of the cross product, the green area vector equals $(1/2) (\vec{a} \times \vec{c}),$ the blue area vector equals $(1/2)(\vec{b} \times \vec{a}),$ the red area vector equals $(1/2)(\vec{c} \times \vec{b})$ and the cyan area vector equals
$$
(1/2)\bigl((\vec{a} - \vec{c}) \times (\vec{b} - \vec{c})\bigr) = (1/2) (\vec{a} \times \vec{b}) - (1/2) (\vec{a} \times \vec{c}) - (1/2) (\vec{c} \times \vec{b}).
$$
Hence the sum of the four area vectors is
\begin{multline*}
(1/2) (\vec{a} \times \vec{c}) + (1/2)(\vec{b} \times \vec{a}) + (1/2)(\vec{c} \times \vec{b}) \\
+ (1/2) (\vec{a} \times \vec{b}) - (1/2) (\vec{a} \times \vec{c}) - (1/2) (\vec{c} \times \vec{b}) = \vec{0}.
\end{multline*}
This proves the theorem for a tetrahedron.
-
In this item I want to explain the connection between the geometric and the coordinate definition of the curl of a vector field.
In the previous item proved that the sum of the area vectors of the four faces of a tetrahedron is the zero vector.
Consider the tetrahedron in the above picture. This tetrahedron has the following properties: its blue face is orthogonal to the coordinate vector $\vec{k}$ (or, equivalently, parallel with $xy$-plane), its green face is orthogonal to the coordinate vector $\vec{\jmath}$ (or, equivalently, parallel with $zx$-plane), its red face is orthogonal to the coordinate vector $\vec{\imath}$ (or, equivalently, parallel with $yz$-plane). Denote by $\vec{n}$ the unit normal vector to the cyan face of the pictured tetrahedron. Denote by $A_r$, $A_g$, $A_b$ and $A_c$ the areas of the red face, green face, blue face and cyan face, respectively. Then we have
\begin{equation*}
- A_r \vec{\imath} -A_g \vec{\jmath} - A_b \vec{k} + A_c \vec{n} = \vec{0}.
\end{equation*}
Consequently,
\begin{equation*}
A_c \vec{n} = A_r \vec{\imath} + A_g \vec{\jmath} + A_b \vec{k} = \bigl\langle A_r, A_g, A_b \bigr\rangle.
\end{equation*}
Our goal is to provide an estimate for the circulation of a smooth vector field \begin{equation*} \vec{F}(x,y,z) = \bigl\langle F_1(x,y,z), F_2(x,y,z), F_3(x,y,z) \bigr\rangle \end{equation*} along the cyan oriented triangle in the picture below. The cyan triangle is in the plane which is orthogonal to the unit vector $\vec{n}$ and it encloses the area of the cyan triangle in the picture above. We will use this estimate to obtain a formula for the circulation per unit area of the vector field $\vec{F}$ at the cyan point $P_c$, which is a fixed point in the plane of $T_c$ as in the picture below. This circulation density is denoted by \begin{equation*} (\operatorname{circ}_{\vec{n}} \vec{F}) (P_c). \end{equation*}
Denote by $T_r$, $T_g$, $T_b$ and $T_c$ the red, green, blue and cyan oriented triangle in the picture below. Looking carefully at the picture below we deduce that \begin{equation*} \oint_{T_c} \vec{F} \cdot d\vec{r} = \oint_{T_r} \vec{F} \cdot d\vec{r} + \oint_{T_g} \vec{F} \cdot d\vec{r} + \oint_{T_b} \vec{F} \cdot d\vec{r}. \end{equation*} Recall: the triangle $T_r$ is parallel to $yz$-plane and it encloses the area $A_r$, the triangle $T_g$ is parallel to $zx$-plane and it encloses the area $A_g$, the triangle $T_b$ is parallel to $xy$-plane and it encloses the area $A_b$ (the areas seen in the picture above).
In the above picture we emphasized four points. Denote the red point which is in plane of the triangle $T_r$ by $P_r$, denote the green point which is in plane of the triangle $T_g$ by $P_g$, denote the blue point which is in plane of the triangle $T_b$ by $P_b$ and denote the cyan point which is in plane of the triangle $T_c$ by $P_c$.
Applying Green's theorem to triangles $T_r$, $T_g$ and $T_b$ we get the following three estimates: \begin{equation*} \oint_{T_r} \vec{F} \cdot d\vec{r} \approx \Bigl( \frac{\partial F_3}{\partial y}(P_r) - \frac{\partial F_2}{\partial z}(P_r) \Bigr) \, A_r, \end{equation*} \begin{equation*} \oint_{T_g} \vec{F} \cdot d\vec{r} \approx \Bigl( \frac{\partial F_1}{\partial z}(P_g) - \frac{\partial F_3}{\partial x}(P_g) \Bigr) \, A_g, \end{equation*} \begin{equation*} \oint_{T_b} \vec{F} \cdot d\vec{r} \approx \Bigl( \frac{\partial F_2}{\partial x}(P_b) - \frac{\partial F_1}{\partial y}(P_b) \Bigr) \, A_b. \end{equation*} Since \begin{equation*} \oint_{T_c} \vec{F} \cdot d\vec{r} = \oint_{T_r} \vec{F} \cdot d\vec{r} + \oint_{T_g} \vec{F} \cdot d\vec{r} + \oint_{T_b} \vec{F} \cdot d\vec{r}, \end{equation*} we have the estimate \begin{equation*} \oint_{T_c} \vec{F} \cdot d\vec{r} \approx \Biggl\langle \frac{\partial F_3}{\partial y}(P_r) - \frac{\partial F_2}{\partial z}(P_r), \frac{\partial F_1}{\partial z}(P_g) - \frac{\partial F_3}{\partial x}(P_g), \frac{\partial F_2}{\partial x}(P_b) - \frac{\partial F_1}{\partial y}(P_b) \Biggr\rangle \cdot \bigl\langle A_r, A_g, A_b \bigr\rangle. \end{equation*} Since \begin{equation*} A_c \vec{n} = \bigl\langle A_r, A_g, A_b \bigr\rangle. \end{equation*} we have \begin{equation*} \frac{1}{A_c} \oint_{T_c} \vec{F} \cdot d\vec{r} \approx \Biggl\langle \frac{\partial F_3}{\partial y}(P_r) - \frac{\partial F_2}{\partial z}(P_r), \frac{\partial F_1}{\partial z}(P_g) - \frac{\partial F_3}{\partial x}(P_g), \frac{\partial F_2}{\partial x}(P_b) - \frac{\partial F_1}{\partial y}(P_b) \Biggr\rangle \cdot \vec{n}. \end{equation*} As triangle $T_c$ collapses to the point $P_c$ the left hand side of the above equation converges to $(\operatorname{circ}_{\vec{n}} \vec{F}) (P_c)$, while the right hand side converges to \begin{equation*} \Biggl\langle \frac{\partial F_3}{\partial y}(P_c) - \frac{\partial F_2}{\partial z}(P_c), \frac{\partial F_1}{\partial z}(P_c) - \frac{\partial F_3}{\partial x}(P_c), \frac{\partial F_2}{\partial x}(P_c) - \frac{\partial F_1}{\partial y}(P_c) \Biggr\rangle \cdot \vec{n}. \end{equation*} Therefore \begin{equation*} (\operatorname{circ}_{\vec{n}} \vec{F}) (P_c) = \Biggl\langle \frac{\partial F_3}{\partial y}(P_c) - \frac{\partial F_2}{\partial z}(P_c), \frac{\partial F_1}{\partial z}(P_c) - \frac{\partial F_3}{\partial x}(P_c), \frac{\partial F_2}{\partial x}(P_c) - \frac{\partial F_1}{\partial y}(P_c) \Biggr\rangle \cdot \vec{n}. \end{equation*} From the above formula it is clear that the circulation density $(\operatorname{circ}_{\vec{n}} \vec{F}) (P_c)$ is the largest in the direction of the vector \begin{equation*} \Biggl\langle \frac{\partial F_3}{\partial y}(P_c) - \frac{\partial F_2}{\partial z}(P_c), \frac{\partial F_1}{\partial z}(P_c) - \frac{\partial F_3}{\partial x}(P_c), \frac{\partial F_2}{\partial x}(P_c) - \frac{\partial F_1}{\partial y}(P_c) \Biggr\rangle. \end{equation*} This vector is called the curl of $\vec{F}$ at the point $P_c$. A useful mnemonic tool is to introduce the nabla operator (see its Wikipedia page) \begin{equation*} \nabla = \Biggl\langle \frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \Biggr\rangle. \end{equation*} With the nabla notation we have \begin{equation*} \operatorname{curl} \vec{F} = \nabla \times \vec{F} = \left| \begin{array}{ccc} \vec{\imath} & \vec{\jmath} & \vec{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ F_1 & F_2 & F_3 \end{array} \right|. \end{equation*} To emphasize that the nabla plays a role of a vector, in class I like adding an arrow over nabla: $\vec{\nabla}$.
- Inspired by the deltoidal torus that we discussed in class and its three-dimensional printout made by Michael Leska I created the animation below which shows that a twisted deltoidal torus can be made of a single piece of long twisted strap.
Place the cursor over the image to start the animation.
- For Section 19.4 do these problems 3, 7, 9, 11, 15, 16, 17, 19, 20, 23, 25, 26, 27, 28, 29, 30, 31.
- For Section 19.3 do these problems 1, 3, 5, 6, 7, 11, 13, 17, 19, 21, 24, 27, 29, 35, 36, 37, 38.
- For Section 21.3 do Exercises 1, 2, 3, 5, 6, 7, 8, 9, 10, 11 (you can solve this problem using vectors as well), 13 (this problem can also be solved using single integrals), 14, 16 , 17, 18, 19, 20.
- Our next topic is the concept of the divergence of a vector field. It is discussed in Section 19.3. The concept of the divergence of the vector field is used in Section 19.4 to establish the Divergence Theorem which provides a connection between the flux of a vector field over a closed positively oriented surface and the triple integral of the divergence of that vector field over the region enclosed by the surface.
- Here is a list of topics for the second exam.
- For Section 19.2 do these problems 7, 10, 11, 14, 17, 25, 26, 29, 31, 38, 39, 41, 43, 45, 46, 47, 48, 51, 53, 55, 59, 61, 63.
- For Section 19.1 do these problems 1, 2, 3, 7, 10, 11, 12, 14, 15, 17, 24, 20, 25, 26, 28, 30, 31, 33, 34, 39, 41, 46, 53, 56, 61, 58, 63, 64, 67, 69, 70.
- During the class today I created this Mathematica file. The file is called 20190730.nb. Right-click on the underlined link; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 8.
- For Section 18.4 do these problems 1, 2, 5, 7, 8, 10, 12, 13, 15, 17, 19, 21, 26, 29, 30, 31, 32, 33, 34, 37, 38, 39, 41, 42.
-
Section 18.4 is a very interesting section. In class I pointed out that whether a vector field is a gradient field or not does depend on its domain. For example, the vector field
\[
\boldsymbol{F}(x,y) = -\frac{y}{x^2+y^2} \boldsymbol{i} + \frac{x}{x^2+y^2} \boldsymbol{j}
\]
is not a gradient field over its natural domain
\[
D_p = \bigl\{(x,y) \in {\mathbb R}^2 : x^2 + y^2 \neq 0 \bigr\} = {\mathbb R}^2 \setminus \bigl\{(0,0)\bigr\}.
\]
Notice that $D_p$ is a punctured plane: the $xy$-plane from which the origin has been removed. To see that $\boldsymbol{F}$ is not a gradient field over this $D_p$, just calculate the circulation of $\boldsymbol{F}$ along the unit circle.
However, if we restrict the domain of $\boldsymbol{F}$ to the cut plane defined as \[ D_c = \bigl\{(x,y) \in {\mathbb R}^2 : y \neq 0 \ \text{or} \ x \gt 0 \bigr\} = {\mathbb R}^2 \setminus \bigl\{(x,0) \in {\mathbb R}^2 \bigl| x \leq 0 \bigr. \bigr\} , \] then the vector field $\boldsymbol{F}$ is a gradient field. In fact, $\boldsymbol{F}$ is the gradient of the function \[ f(x,y) = \begin{cases} -\arctan(x/y) + \pi/2 & \quad \text{if} \quad y \gt 0, \\ \phantom{-}0 & \quad \text{if} \quad y = 0 \quad \text{and} \quad x \gt 0 , \\ -\arctan(x/y) - \pi/2 & \quad \text{if} \quad y \lt 0. \end{cases} \] Notice that the domain of the function $f$ is exactly the set $D_c.$ Also, notice that $D_c$ is the $xy$-plane from which the origin and the negative $x$-half-axis have been removed. -
The formula for $f(x,y)$ above looks complicated but this is not too complicated function. Here is its plot.
- Problem 33 in Section 18.4 is an interesting problem. This problem deals with the curve called the Folium of Decartes. As pointed out in the book this curve is given by the implicit equation \[ x^3+y^3 - 3 x y =0. \] The vector equation for this curve is \[ \left\langle \frac{3t}{1+t^3}, \frac{3t^2}{1+t^3} \right\rangle, \qquad t \in (-\infty,-1) \cup (-1,+\infty) = \mathbb{R}\setminus\{-1\}. \]
- In this item I explain one way to understand the graph of the Folium of Decartes without using the technology. It is not directly related to the question. We first figure out the intersections of the Folium of Decartes with the lines $y = c - x.$ So, substitute $y = c -x$ in the implicit equation $x^3+y^3 - 3 x y =0.$ The resulting equation is a quadratic equation in $x$ with $c$ as a parameter. The following interesting trichotomy holds:
- For $c = 0$ and $c = 3$ we have a unique solution for $x.$ So we found that the points $(0,0)$ and $(3/2,3/2)$ are on the curve. These are the black and the red point on the picture below.
- For $c \in (-1,0) \cup (0,3)$ we have two distinct solutions for $x.$ These solutions result in two points on the curve which are symmetric with respect to the line $y=x$ and lie on the line $y=c-x$ symmetric with respect to the point $(c/2,c/2).$ A pair of such points are the points $\bigl(2^{2/3},2^{2/3}\bigr)$ and $\bigl(2^{2/3},2^{2/3}\bigr).$ The first of these points is maroon and the second is light red on the picture below. For a positive $c$ the corresponding points lie in the first quadrant, one above, the other below the diagonal. For negative $c$ one of the points lies in the second and the symmetric one in the fourth quadrant.
- For $c \in (-\infty,-1]\cup (3, + \infty)$ there are no solutions for $x.$ That is, for such $c$-s the lines $y=c-x$ do not intersect the given curve. The significance of the line $y = -1 - x$ is that it is an asymptote for the Folium of Decartes.
-
With the above analysis of the Folium of Decartes we understand the plot below better.
On the plot above I emphasised the part of the curve corresponding to $t \in (0,1)$ in dark blue. That is the part between the black point and the red point which is below the diagonal and contains the maroon point. The part between the red and black point which is above the diagonal is in blue. This part contains the light red point. On the curve I indicated the direction of the increasing $t$ by arrows.
- The textbook asks us to calculate the area in the first quadrant enclosed by the Folium of Decartes. The part of the curve in the first quadrant corresponds to the parameter $t \in [0,+\infty).$ To calculate the area we need to evaluate the following line integral \begin{align*} \oint_C x \boldsymbol{j} \cdot d\boldsymbol{r} & = \int_0^{\infty} \left\langle 0, \frac{3t}{1+t^3} \right\rangle \cdot \left\langle \frac{3-6 t^3}{\left(t^3+1\right)^2}, \frac{3 t \left(2 - t^3\right)}{\left(t^3+1\right)^2} \right\rangle dt \\ & = \int_0^{\infty} \frac{9 t^2 \left(2 - t^3 \right)}{\left(t^3+1\right)^3} dt \\ & = \int_0^{\infty} \frac{18 t^2 }{\left(t^3+1\right)^3} dt - \int_0^{\infty} \frac{9t^5}{\left(t^3+1\right)^3} dt \\ & = \int_1^{\infty} \frac{6}{u^3} du - \int_1^{\infty} \frac{3(u-1)}{u^3} du \\ & = 9 \int_1^{\infty} \frac{1}{u^3} du - 3 \int_1^{\infty} \frac{1}{u^2} du \\ & = -\frac{9}{2} \left. \frac{1}{u^2} \right|_1^{\infty} + 3 \left. \frac{1}{u} \right|_1^{\infty} \\ & = \frac{9}{2} - 3 \\ &= \frac{3}{2} \end{align*}
- In the above calculation we encountered an improper integral. One is justified in being a little suspicions about the validity of that method. Luckily, in this problem we can avoid the improper integral by calculating one-half of the asked area which is below the diagonal. To accomplish this we need to calculate the line integral of the same vector field along the yellow oriented line segment in the picture above. This line integral evaluates as follows \[ \int_0^{3/2} \left\langle 0, \frac{3}{2} - s \right\rangle \cdot \left\langle -1,-1 \right\rangle ds = \int_0^{3/2} \left( s - \frac{3}{2} \right) ds = -\frac{9}{8} \] Now we calculate the line integral along the dark blue part of the curve. This part of the curve corresponds to the parameter $t \in [0,1].$ We calculate \begin{align*} \int_{C_b} x \boldsymbol{j} \cdot d\boldsymbol{r} & = \int_0^{1} \left\langle 0, \frac{3t}{1+t^3} \right\rangle \cdot \left\langle \frac{3-6 t^3}{\left(t^3+1\right)^2}, \frac{3 t \left(2 - t^3\right)}{\left(t^3+1\right)^2} \right\rangle dt \\ & = \int_0^{1} \frac{18 t^2 }{\left(t^3+1\right)^3} dt - \int_0^{1} \frac{9t^5}{\left(t^3+1\right)^3} dt \\ & = \int_1^{2} \frac{6}{u^3} du - \int_1^{2} \frac{3(u-1)}{u^3} du \\ & = -\frac{9}{2} \left. \frac{1}{u^2} \right|_1^{2} + 3 \left. \frac{1}{u} \right|_1^{2} \\ & = - \frac{9}{8} + \frac{9}{2} + \frac{3}{2} - 3 \\ &= \frac{15}{8} \end{align*} Thus, the area is $2(15/8 - 9/8) = 3/2.$
-
Problem 29 in Section 18.4 is also an interesting problem. Two oriented curves are given in this problem. In the plot below the given curve $C_1$ is the purple curve and the given curve $C_2$ is the navy blue curve.
In part (b) of this problem we are asked to calculate the line integral of the vector field
\begin{equation*}
\boldsymbol{F} = \bigl\langle yz, z x + z, xy+y+1 \bigr\rangle
\end{equation*}
along the navy blue curve. It turns out that calculating this line integral is difficult. To simplify the calculation we observe that the curl of the vector field $\boldsymbol{F}$ is $\boldsymbol{0}.$ Since the vector field $\boldsymbol{F}$ is everywhere defined the curl test tells us that $\boldsymbol{F}$ is a gradient field. Therefore $\boldsymbol{F}$ is path-independent. Consequently, the line integral along the navy blue curve equals the sum of the line integral along the purple curve plus the line integral along the olive curve. Both of these integrals are easy. The line integral along the purple curve is
\begin{align*}
\int_0^2 \bigl\langle 0, & t^2 (\cos 2 \pi t) (\sin 2\pi t) + t (\sin 2\pi t), 1 \bigr\rangle \\
& \cdot \bigl\langle (\cos 2 \pi t) - 2 \pi t (\sin 2 \pi t) , 0, (\sin 2\pi t) + 2 \pi t (\cos 2\pi t) \bigr\rangle dt \\
& = \int_0^2 \bigl( (\sin 2\pi t) + 2 \pi t (\cos 2\pi t) \bigr) dt \\
& = \Bigl( t (\sin 2\pi t) \Bigr) \Bigl. \Bigr|_0^2 \\
& = 0.
\end{align*}
The line integral along the olive curve is
\[
\int_0^2 \bigl\langle 0, 0, 2 t + t + 1 \bigr\rangle \cdot \bigl\langle 0,1,0 \bigr\rangle dt = \int_0^2 0 dt = 0.
\]
Another way to calculate the requested line integral is to observe that \begin{equation*} \boldsymbol{F} = \bigl\langle yz, z x + z, xy+y+1 \bigr\rangle = \operatorname{grad} \bigl( xyz + yz + z \bigr) \end{equation*} and calculate the value of $f(x,y,z) = xyz + yz + z$ at the red point $A = (0,0,0)$ and the red point $B = (2,2,0)$. Then the requested line integral equals $f(B) - f(A) = 0.$
- For Section 18.3 do these problems 1, 3, 4, 5, 7, 8, 9, 13, 15, 17, 19, 20, 23, 26, 28, 29, 31, 33, 34, 37, 49, 53, 54, 57.
- Since we are working with vectors in this class, and since I am using colors to clarify relationship between various mathematical quantities that we encounter, I hope you will find it useful to see how colors can be organized using vectors in the unit cube; see the web-page dedicated to the Color Cube
- Today we calculated line integrals involving the cardioid. We noticed how symmetries can be used to simplified calculations. Specifically we calculated \[ \int_C \boldsymbol F \cdot d \boldsymbol r = 4 \] where \[ \boldsymbol F(x,y) = \langle x, y \rangle \] and $C$ is given by \[ \boldsymbol r(t) = \bigl\langle 2 (\cos t) - \cos(2t) ,2 (\sin t) - \sin(2t) \bigr\rangle, \quad 0 \leq t \leq \pi. \]
-
The animation below illustrates the process of calculating the line integral of the navy blue force along the maroon curve. In the animation we show the whole cardioid, not just half of it as given by the formula.
Place the cursor over the image to start the animation.
- To calculate a given line integral it is essential to calculate the dot product of the force at a point on the given curve and the tangent vector at the same point of the curve. This is the dot product of the navy blue vector and the maroon vector in the above animation. Using the given formulas this dot product calculates as \[ \boldsymbol F(\boldsymbol r(t)) \cdot \boldsymbol r'(t) = 2 \sin t, \quad 0 \leq t \leq \pi. \] This function reveals all the symmetries in the force and the curve. The quantity $(2 \sin t) \Delta t$ with small positive $\Delta t$ is an approximation of the work done by the navy blue force $\boldsymbol F$ near the maroon end-point of the vector $\boldsymbol r(t)$ which is not pictured in the above animation.
- It is also interesting to calculate line integrals along the cardioid with the rotational vector field: \[ \int_C \boldsymbol F \cdot d \boldsymbol r = 12 \pi \] with \[ \boldsymbol F(x,y) = \langle -y,x \rangle \] and where $C$ is given by \[ \boldsymbol r(t) = \bigl\langle 2 (\cos t) - \cos(2t) ,2 (\sin t) - \sin(2t) \bigr\rangle, \quad 0 \leq t \leq 2\pi. \]
-
The animation below illustrates the process of calculating the line integral of the navy blue force along the maroon curve.
Place the cursor over the image to start the animation.
- To calculate a given line integral it is essential to calculate the dot product of the force at a point on the given curve and the tangent vector at the same point of the curve. This is the dot product of the navy blue vector and the maroon vector in the above animation. Using the given formulas this dot product calculates as \[ \boldsymbol F(\boldsymbol r(t)) \cdot \boldsymbol r'(t) = 6-6 \cos t, \quad 0 \leq t \leq 2\pi. \] This function reveals all the symmetries in the force and the curve. The quantity $(6-6 \cos t) \Delta t$ with small positive $\Delta t$ is an approximation of the work done by the navy blue force $\boldsymbol F$ near the maroon end-point of the vector $\boldsymbol r(t).$
- For Section 18.1 do these problems 1-6, 7, 9, 13, 14, 15, 17, 19, 21, 22, 25, 27, 28-31, 32, 33, 35, 42-46, 47, 49, 51, 55, 56, 57, 58.
- For Section 18.2 do these problems 1-3, 5, 7, 8, 13, 14, 17, 19, 20, 21, 22, 23, 24, 26, 28, 30, 31, 32, 35, 36, 38, 39, 41.
- Today I handed out the extended version of Assignment 1. The assignment is due next Thursday, July 25.
- For Section 17.3 do these problems 1-6, 7, 10, 12, 13, 15, 17, 21, 22, 24, 25, 27, 28, 29, 30, 31, 32-34, 38, 39.
- For Section 17.4 do these problems 1, 3, 4-9, 16, 19, 20, 23, 24. In this section pay special attention to Examples 1, 2 and 3.
- In class I did a problem similar to Example 2: Find the parametric equations $\bigl(x(t),y(t)\bigr)$ of the flow line of the vector field $\boldsymbol{F}(x,y) = y \boldsymbol{i} + \boldsymbol{j}$ such that $\bigl(x(0),y(0)\bigr) = (0,-1).$ Also find the implicit equation of this flow line. I might have made a mistake in class. Please redo this problem carefully.
- Do this problem: Find the parametric equations $\bigl(x(t),y(t)\bigr)$ of the flow line of the vector field $\boldsymbol{F}(x,y) = \boldsymbol{i} + y \boldsymbol{j}$ such that $\bigl(x(0),y(0)\bigr) = (0,-2).$ Also find the implicit equation of this flow line.
- For Section 21.2 do Exercises 1, 3, 4, 5, 6, 9, 10, 11, 13, 14, 15, 16.
-
Below I list few additional problems in which change of coordinates in double integrals is quite helpful.
- Evaluate the finite area in the first quadrant bounded by the graphs of the following functions: $y=1/(3x)$, $y = 1/x$, $y = 2x$, $y = x/2$.
- Let $n$ be a positive integer greater than $1$. Denote by $A_n$ the finite area in the first quadrant bounded by the graphs of the following functions: $y=1/(nx)$, $y = 1/x$, $y = 2x$, $y = x/2$. Find the following limit $\lim_{n\to\infty} A_n.$ Verify your answer by calculating two single integrals.
- Evaluate the double integral $\iint_R \exp\left(\frac{x-y}{x+y}\right) dA$ where the region $R$ is the triangle with the vertices $(0,0),$ $(1,0),$ $(0,1).$ This integral is in some sense improper integral since the function $\exp\left(\frac{x-y}{x+y}\right)$ is not defined at $(0,0).$ It turns out that this is not a problem in evaluation. However, if one wants to be rigorous, one can calculate the integral over the trapezoid with vertices $(0,a),$ $(a,0),$ $(1,0),$ $(0,1),$ where $a \gt 0,$ and then take the limit as $a \to 0.$
- Let $n$ be a positive integer. Evaluate the double integral $\iint_R (x+y)^n dA$ where the region $R$ is the triangle with the vertices $(0,0),$ $(1,0),$ $(0,1).$
- Let $n$ be a positive integer. Evaluate the double integral $\iint_R (x-y)^n dA$ where the region $R$ is the triangle with the vertices $(0,0),$ $(1,0),$ $(0,1).$
- Let $a$ and $b$ be positive real numbers such that $a \lt b.$ Let $S_a$ be the sphere with radius $a$ and let $S_b$ be the sphere with radius $b$ both centered at the origin. Evaluate the triple integral $\iiint_W (x^2+y^2+z^2)^{-3/2} dV$ where $W$ is the region inside the sphere $S_b$ and outside the sphere $S_a.$
- The exam will cover 16.1 through 16.5, 17.1, 17.2, 21.1 and 21.2.
- Today we continued with Section 21.1. Do these problems 1-4, 5-8, 9-12, 13, 15, 18, 19, 22, 23, 26, 27, 31, 32, 33, 34, 35, 36.
- During the class today I created this Mathematica file. The file is called 20190710.nb. Right-click on the underlined link; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 8.
-
In class we derived the following parametric equations of the torus
\[
\begin{array}{rl}
x = u(s,t) \!\!& = (b + a \sin s) ( \cos t), \\
y = v(s,t) \!\!& = (b + a \sin s) ( \sin t), \\
z = w(s,t) \!\!& = a \cos s, \\
\text{where} & s, t \in [0,2 \pi].
\end{array}
\]
Here we assume that $a \lt b.$ This torus is obtained as a circle of radius $a$ rotates about $z$-axis with the radius $b.$
Whenever we encounter a parametric equations of a surface it is important to identify so called parameter curves. We obtain parameter curves for the torus by keeping one parameter constant.
Set $s=s_0,$ where $s_0 \in [0,2 \pi],$ in the above parametric equations of the torus. Then we obtain the curve \[ \begin{array}{rl} x = u(s_0,t) \!\!& = (b + a \sin s_0) ( \cos t), \\ y = v(s_0,t) \!\!& = (b + a \sin s_0) ( \sin t), \\ z = w(s_0,t) \!\!& = a \cos s_0, \\ \text{where} & t \in [0,2 \pi]. \end{array} \] This is the circle in the horizontal plane $z = a \cos s_0$ centered at the point $(0,0,a \cos s_0)$ with the radius $b + a \sin s_0.$ The four most distinguished circles of this kind are as follows: For $s_0 = 0$ we obtain the highest circle at $z = a$ whose radius is $b$; for $s_0 = \pi$ we obtain the lowest circle at $z = -a$ whose radius is also $b$; $s_0 = \pi/2$ we obtain the largest circle of this kind at $z = 0$ whose radius is $b+a$; $s_0 = 3\pi/2$ we obtain the smallest circle of this kind also in the plane $z = 0$ whose radius is $b-a$. The animation below shows 48 such circles colored purple.Place the cursor over the image to start the animation.
Now set $t=t_0,$ where $t_0 \in [0,2 \pi],$ in the above parametric equations of the torus. Then we obtain the curve \[ \begin{array}{rl} x = u(s,t_0) \!\!& = (b + a \sin s) ( \cos t_0), \\ y = v(s,t_0) \!\!& = (b + a \sin s) ( \sin t_0), \\ z = w(s,t_0) \!\!& = a \cos s, \\ \text{where} & s \in [0,2 \pi]. \end{array} \] It is easier to look at the vector equation of this curve: \[ (b + a \sin s) ( \cos t_0) \boldsymbol{i} + (b + a \sin s) ( \sin t_0)\boldsymbol{j} + (a \cos s) \boldsymbol{k} \] which after grouping the terms becomes \[ b\bigl( (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr) + a \Bigl( (\cos s) \boldsymbol{k} + (\sin s) \bigl( (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr) \Bigr). \] To understand this equation, first notice that \[ ( \cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \] is a constant unit vector in $xy$-plane. Therefore \[ (\cos s) \boldsymbol{k} + (\sin s) \bigl( ( \cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr), \quad s \in [0,2\pi], \] is a unit circle centered at $(0,0,0)$ which is vertical to $xy$-plane. The special cases of this unit circle are when $t_0 = 0$ we have \[ (\cos s) \boldsymbol{k} + (\sin s) \boldsymbol{i}, \quad s \in [0,2\pi]; \] for $t_0 = \pi/2$ we have \[ (\cos s) \boldsymbol{k} + (\sin s) \boldsymbol{j}, \quad s \in [0,2\pi]; \] for $t_0 = \pi$ we have \[ (\cos s) \boldsymbol{k} - (\sin s) \boldsymbol{i}, \quad s \in [0,2\pi]; \] for $t_0 = 3\pi/2$ we have \[ (\cos s) \boldsymbol{k} - (\sin s) \boldsymbol{j}, \quad s \in [0,2\pi]. \] The next step is to observe that \[ a \Bigl( (\cos s) \boldsymbol{k} + (\sin s) \bigl( ( \cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr) \Bigr), \quad s \in [0,2\pi], \] is a circle of radius $a$ centered at $(0,0,0)$ and finally the vector $b\bigl( (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr)$ moves the center of this circle to the point $(\cos t_0,\sin t_0, 0)$. Thus \[ b\bigl( (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr) + a \Bigl( (\cos s) \boldsymbol{k} + (\sin s) \bigl( (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j} \bigr) \Bigr). \] is the circle of radius $a$ centered at the point $(\cos t_0,\sin t_0, 0)$ and in the plane spanned by the vectors $\boldsymbol{k}$ and $ (\cos t_0) \boldsymbol{i} + (\sin t_0)\boldsymbol{j}.$ The animation below shows 48 such circles colored teal.Place the cursor over the image to start the animation.
In the animation below I show how a torus can be obtained from a sphere with radius $a.$ First we move all the meridians of the sphere away from the $z$-axis to the distance $b.$ Then we extend the meridians which are half-circles to full-circles with radius $a.$
Place the cursor over the image to start the animation.
- For Section 21.1 do these problems 1-4, 5-8, 9-12, 13, 15, 18, 19, 22, 23, 26, 27, 31, 32, 33, 34, 35, 36.
- For Section 17.2 do these problems 1, 3, 5, 7, 8, 10, 12, 13, 15, 15, 17, 19, 21, 22, 23, 27, 28, 29, 30, 32, 36, 37, 41, 42, 44.
-
During the class today I created
this Mathematica file. The file is called 20190709.nb. Right-click on the underlined link; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 8. You will find Mathematica 8 on computers in BH 215 and in the Math Center. Look in
Start -> Wolfram Mathematica -> Mathematica 8. Open Mathematica first, then open 20190405.nb from Mathematica. - When you open the file 20190709.nb you can go through the notebook cell by cell and evaluate each cell individually, or you can evaluate the entire notebook by the keyboard sequence Alt+v o or with the manu item Evaluate ⇾ Evaluate Notebook.
- More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
-
As an example of a surface of revolution I derived parametric equations of a nice vase produced by rotating the graph of $y = 2+\sin x$, $x \in [0,2 \pi]$, about the $x$-axis. The animation below illustrates how revolution creates a vase.
Place the cursor over the image to start the animation.
-
Below I list few problems in which arc length of a curve can be calculated explicitly.
- Problem 1. Calculate the arc length of the graph of the function $y = x^{3/2}$ between the points $(0,0)$ and $(1,1)$.
- Problem 2. Calculate the arc length of the graph of the function $y= (1/4) x^2-(1/2) \ln x$, between the points $(1,1/4)$ and $\bigl(e,(e^2-2)/4\bigr)$.
- Problem 3. Calculate the arc length of the cycloid given by the parametric equations \[ x(t) = t- \sin t , \quad y(t) = 1-\cos t \quad \text{where} \quad 0 \leq t \leq 2 \pi. \]
- Problem 4. Calculate the arc length of the astroid curve given by the parametric equations \[ x(t) = (\cos t)^3, \quad y(t) = (\sin t)^3 \quad \text{where} \quad 0 \leq t \leq 2 \pi. \]
- Problem 5. Calculate the arc length of the spiral given by the parametric equations \[ x(t) = (\exp t)(\cos t), \quad y(t) = (\exp t)(\sin t) \quad \text{where} \quad -\pi \leq t \leq \pi. \]
- Problem 6. Calculate the arc length of the cardioid given by the parametric equations \[ x(t) = (1+\cos t)\cos t, \quad y(t) = (1+\cos t) \sin t \quad \text{where} \quad 0 \leq t \leq 2\pi. \]
-
You will probably notice that what I call cardioid in Problem 6 differs from the cardioid that we derived in class. The parametric equations that we deduced for a cardioid in class were
\[
x = u_1(t) = 2 \cos(t) -\cos(2t), \quad y = v_1(t) = 2 \sin(t) -\sin(2t).
\]
The cardioid that we derived in class is the cardioid in Problem 6 scaled by two and translated to the point $(1,0)$. This is justified by the following calculations:
\begin{align*}
u_1(t) & = 2 \cos(t) -\cos(2t) \\
& = 2 \cos(t) -(\cos t)^2 +(\sin t)^2 - (\cos t)^2 - (\sin t)^2 + 1 \\
& = 2 \cos(t) - 2(\cos t)^2 + 1 \\
& = 2(1-\cos t) \cos t + 1 \\
& = 2 u(t) + 1, \\
v_1(t) & = 2 \sin(t) -\sin(2t) \\
& = 2 \sin(t) - 2 (\sin t) (\cos t)\\
& = 2(1-\cos t) \sin t \\
& = 2 v(t).
\end{align*}
Hence
\[
\bigl(u_1(t),v_1(t)\bigr) = 2\bigl(u(t),v(t)\bigr) + (1,0).
\]
Or, in pictures,
The cardioid from Problem 6
The cardioid from Problem 6 scaled by 2
The cardioid from Problem 6 scaled by 2 and translated.
- For Section 17.2 do these problems 1, 3, 5, 7, 8, 10, 12, 13, 15, 15, 17, 19, 21, 22, 23, 27, 28, 29, 30, 32, 36, 37, 41, 42, 44.
-
Here is a derivation of the equation of a cardioid.
Click through several helpful images.
- The origin $O = (0,0)$ of the coordinate system is the gray point. Denote by $C$ the black point in the picture.
- Both circles in the picture are unit circles, the gray one centered at $O,$ the black one centered at $C.$
- The angle at the gray point marked by the blue arc between the green and the blue point measures $t$ radians. That is, the length of this blue arc is $t.$
- There are two other angles in the picture which also measure $t$ radians: the angle marked by the blue arc between the blue and the red point and the angle marked by the black arc between the blue point and the purple point.
- The green point is $(1,0),$ the blue point is $(\cos t, \sin t),$ the black point $C$ is $(2\cos t, 2\sin t),$ the purple point is $(2\cos t-1, 2\sin t).$
- The goal is to determine the coordinates of the red point.
- Denote by $\boldsymbol{a}$ the unit vector whose tail is the black point and whose head is the red point. Based on the previous items we have \begin{align*} \boldsymbol{a} & = \bigl(\cos( \pi+2 t ) \bigr) \boldsymbol{i} + \bigl(\sin( \pi+2 t ) \bigr) \boldsymbol{j} \\ & = -\bigl(\cos(2 t ) \bigr) \boldsymbol{i} - \bigl(\sin(2 t ) \bigr) \boldsymbol{j}. \end{align*}
- Denote by $\boldsymbol{c}$ the position vector of the black point. Then the position vector of the red point is \[ \boldsymbol{c} + \boldsymbol{a} = \bigl( 2 \cos t - \cos(2 t ) \bigr) \boldsymbol{i} + \bigl(2 \sin t - \sin(2 t ) \bigr) \boldsymbol{j}. \]
- Thus the parametric equations for this cardioid are \begin{align*} x & = 2 \cos t - \cos(2 t ) \\ y & = 2 \sin t - \sin(2 t ) \end{align*}
-
The parametrization presented in the previous item is implemented below to animate one unit circle roling around a fixed unit circle.
Place the cursor over the image to start the animation.
- As I explained in class, similar reasoning can be used to derive that the parametric equations of the curve obtained when a wheel of radius $a$ rolls along the unit circle: For $0 \leq t \leq 2 \pi$, \begin{align*} x & = (1+a) \cos t - a \cos\bigl( (1+1/a) t \bigr) \\ y & = (1+a) \sin t - a \sin\bigl( (1+1/a) t \bigr) \end{align*} This curve is called a epicycloid. Below I show few special cases.
- When the radius of the rolling is $a=1/2$ the curve is called a nephroid. This name means "kidney-shaped" \begin{align*} x & = \frac{3}{2} \cos t - \frac{1}{2} \cos\bigl( 3 t \bigr) = 3 \cos t - 2 (\cos t)^3 \\ y & = \frac{3}{2} \sin t - \frac{1}{2} \sin\bigl( 3 t \bigr) = 2 (\sin t)^3 \end{align*}
- The same reasoning used in the derivation of the parametric equations for the cardioid leads to the parametric equations of curves obtained by a small circle rolling inside the unit circle. The parametric equations of the curve obtained when a wheel of radius $a$ rolls inside the unit circle are: For $0 \leq t \leq 2 \pi$, \begin{align*} x & = (1-a) \cos t + a \cos\bigl( (1-1/a) t \bigr) \\ y & = (1-a) \sin t + a \sin\bigl( (1-1/a) t \bigr) \end{align*} This curve is called hypocycloid. Below I show few special cases.
- When the radius of the circle rolling inside the unit circle is $a=1/3$ the curve is called an deltoid, see also deltoid at MathWorld . \begin{align*} x & = \frac{2}{3} \cos t + \frac{1}{3} \cos\bigl(2 t \bigr) \\ y & = \frac{2}{3} \sin t - \frac{1}{3} \sin\bigl(2 t \bigr) \end{align*}
- When the radius of the circle rolling inside the unit circle is $a=1/4$ the curve is called an astroid. \begin{align*} x & = \frac{3}{4} \cos t + \frac{1}{4} \cos\bigl( 3 t \bigr) \\ y & = \frac{3}{4} \sin t - \frac{1}{4} \sin\bigl( 3 t \bigr) \end{align*}
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
- For Section 17.1 do these problems 1, 2, 4, 5, 6, 11, 12, 14, 17, 18, 19, 20, 21, 23, 26, 27, 30, 23, 35, 39, 40, 31, 42, 43-48, 52-62, 74, 76, 81, 84, 89, 95.
-
In this item I will illustrate the derivation of the parametric equations for the cycloid.
Place the cursor over the image to start the animation.
The following animation the moving circle is pictured as a bicycle wheel.Place the cursor over the image to start the animation.
- The most important concept for double integrals in polar coordinates is the area element in polar coordinates: $dA = r \, d\theta \, d r$.
-
The first step to understanding the spherical coordinates is the understanding of the unit vectors in $\mathbb{R}^3.$
- The importance of the trigonometric functions $\cos$ and $\sin$ is that each point on the unit circle in $\mathbb{R}^2$ can be expressed as $\bigl( \cos t, \sin t \bigr)$ with $t \in [0, 2\pi)$, or with $t\in (-\pi,\pi].$ This is the reason that trigonometric functions are also called circular functions.
- If we denote the coordinate unit vectors in $\mathbb{R}^2$ by $\boldsymbol{i}$ and $\boldsymbol{j},$ then each unit vector $\boldsymbol{u}$ in $\mathbb{R}^2$ can be expressed as \[ \boldsymbol{u} = (\cos \theta) \boldsymbol{i} + (\sin \theta) \boldsymbol{j}, \qquad \quad \theta \in [0,2 \pi). \]
-
Let now $\boldsymbol{v}$ be a unit vector in the three-dimensional space $\mathbb{R}^3.$ That is
\[
\boldsymbol{v} = v_1 \boldsymbol{i} + v_2 \boldsymbol{j} + v_3 \boldsymbol{k}.
\]
In the figure below the choice of colors is not random:
- $\boldsymbol{v}$ is the black vector;
- $\boldsymbol{i}$ is the red vector;
- $\boldsymbol{j}$ is the green vector;
- $\boldsymbol{k}$ is the blue vector;
- the yellow vector is the unit vector in the direction of the projection of $\boldsymbol{v}$ onto $xy$-plane. Say, as before, \[ \boldsymbol{u} = (\cos \theta) \boldsymbol{i} + (\sin \theta) \boldsymbol{j}, \qquad \quad \theta \in [0,2 \pi); \] The angle $\theta$ is also colored yellow.
- $\phi \in (0,\pi)$ is the gray angle between $\boldsymbol{v}$ and $\boldsymbol{k}$.
Click on the image to animate in $\theta$; double-click to animate in $\phi$.
- In conclusion, each unit vector $\boldsymbol{v}$ in three-dimensional space is uniquely determined by two angles: the angle $\phi \in [0,\pi]$ between $\boldsymbol{v}$ and $\boldsymbol{k}$ and the angle $\theta \in [0,2 \pi)$ which is the angle between the projection of $\boldsymbol{v}$ onto $xy$-plane and $\boldsymbol{i}$. Then \[ \boldsymbol{v} = (\sin \phi)(\cos \theta) \boldsymbol{i} + (\sin \phi)(\sin \theta) \boldsymbol{j} + (\cos \phi) \boldsymbol{k}. \]
-
To enrich our toolbox for calculating triple integrals we need to understand cylindrical and spherical coordinates. The most important concepts here are the volume element in cylindrical coordinates: $dV = r \, d\theta \, d r \, dz$ and
the volume element in spherical coordinates is: $dV = \rho^2 (\sin \phi) \, d\theta\, d\phi \, d\rho$. I created the image below to help you understand this formula.
-
In the problems below I use the following notation. The unit disk in $\mathbb{R}^2$ is denoted by $\mathbb{D}$. That is
\[
\mathbb{D} = \bigl\{ (x,y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1 \bigr\}.
\]
The special square in $\mathbb{R}^2$ whose edges are tangent to $ \mathbb{D}$ and parallel to the coordinate axes is denoted by $\mathbb{Q}$. That is,
\[
\mathbb{Q} = \bigl\{ (x,y) \in \mathbb{R}^2 | x,y \in [-1,1] \bigr\}.
\]
The unit ball in $\mathbb{R}^3$ is denoted by $\mathbb{B}$. That is
\[
\mathbb{B} = \bigl\{ (x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 \leq 1 \bigr\}.
\]
The special cube in $\mathbb{R}^3$ whose sides are tangent to $\mathbb{B}$ and parallel to the coordinate planes is denoted by $\mathbb{U}$. That is,
\[
\mathbb{U} = \bigl\{ (x,y,z) \in \mathbb{R}^2 | x,y,z \in [-1,1] \bigr\}.
\]
- Problem 1. Find the average distance of a point in $\mathbb{D}$ to the center $(0,0)$ of $\mathbb{D}.$
- Problem 2. Find the average distance of a point in $\mathbb{D}$ to a fixed diameter of $\mathbb{D}.$ Here you can choose a diameter along the $x$-axis or along the $y$-axis.
- Problem 3. Find the average distance of a point in $\mathbb{D}$ to the coordinate axes. Recall that the distance of a point $(x,y)$ to the coordinate axes is given by $\min\{|x|,|y|\}.$
- Problem 4. Find the average distance of a point in $\mathbb{Q}$ to the $x$-axis.
- Problem 5. Find the average distance of a point in $\mathbb{Q}$ to a fixed diagonal of $\mathbb{Q}.$ Here you can choose the diagonal joining the points $(-1,-1)$ and $(1,1).$
- Problem 6. Find the average distance of a point in $\mathbb{Q}$ to the coordinate axes.
- Problem 7. Find the average distance of a point in $\mathbb{B}$ to the center $(0,0,0)$ of $\mathbb{B}.$
- Problem 8. Find the average distance of a point in $\mathbb{B}$ to a fixed diameter of $\mathbb{B}.$ Here you can chose the diameter along the $z$-axis.
- Problem 9. Find the average distance of a point in $\mathbb{B}$ to $xy$-plane.
- Problem 10. Find the average distance of a point in $\mathbb{B}$ to the coordinate planes. Recall that the distance of a point $(x,y,z)$ to the coordinate planes is given by $\min\{|x|,|y|,|z|\}.$ (This is the hardest problem on this list.)
- Problem 11. Find the average distance of a point in $\mathbb{U}$ to $xy$-plane.
- Problem 12. Find the average distance of a point in $\mathbb{U}$ to the coordinate planes.
-
In this item we want to calculate
$$
\iint_R \sqrt{xy} dA
$$
where
$$
R = \bigl\{ (x,y) \in {\mathbb R}\times{\mathbb R} \, : \, 0 \leq x \leq 1, \ 0 \leq y \leq 1-x \bigr\}.
$$
- First I present an animated gif file illustrating iterated integration of the function $\sqrt{xy}$ over the region $$ R = \bigl\{ (x,y) \in {\mathbb R}\times{\mathbb R} \, : \, 0 \leq x \leq 1, \ 0 \leq y \leq 1-x \bigr\}. $$
- In the above animation for a fixed $x\in[0,1]$ in navy blue we emphasise the slice under the graph of $\sqrt{xy}$ with the values of $y$ in the closed interval $[0,1-x].$ The area of the navy blue slice is $$ \int_0^{1-x} \sqrt{xy} \, dy = \frac{2}{3} \sqrt{x} (1-x)^{3/2}. $$
- To get the volume of the tent in the above picture we would slice the volume of the tent into $m$ vertical slices parallel to the $yz$-plane and of the thickness $\Delta x.$ Then the approximate volume of the tent would be $$ \sum_{i=1}^m \frac{2}{3} \sqrt{x_i} (1-x_1)^{3/2} \, \Delta x. $$ The above sum is a Riemann sum of the definite integral $$ \frac{2}{3} \int_0^{1} \sqrt{x} (1-x)^{3/2} dx. $$ Therefore the volume of the tent is the above definite integral.
- Thus, to calculate the volume of the tent we need to evaluate the integral $$ I = \int_0^{1} (1-x) \sqrt{x(1-x)} \, dx. $$ Since the function $y= (1-x) \sqrt{x(1-x)}$ is symmetric with respect to the line $x = 1/2$ we have $$ I = \int_0^{1} x \sqrt{x(1-x)} \, dx. $$ (To conclude that the preceding two integrals evaluate to the same number one can also introduce the change of variables $u=1-x.$) Adding the preceding two equalities yields $$ 2 I = \int_0^{1} \sqrt{x(1-x)} \, dx. $$
- To evaluate the last integral we observe that $y = \sqrt{x(1-x)}$ is the top part of the circle of radius $1/2$ centered at $(1/2,0)$. This claim is justified by doing algebra: for $y \gt 0$ the equality $y = \sqrt{x(1-x)}$ is equivalent to $y^2 = x-x^2$, which is furter equivalent to $(x-1/2)^2 + y^2 = 1/4.$ Consequently, $$ 2 I = \int_0^{1} \sqrt{x(1-x)} \, dx = \frac{\pi}{8}. $$
- Finally we conclude that the volume of the pictured tent is $$ \frac{2}{3} \int_0^{1} \sqrt{x} (1-x)^{3/2} dx = \frac{2}{3} \frac{\pi}{16} = \frac{\pi}{24} . $$
- Is this value reasonable? Compare this tent to the pyramid with the same base and whose apex is at the highest point of the tent which is $(1/2,1/2,1/2).$ The volume ot that pyramid is $1/12 = 2/24.$ As expected it is smaller than the volume of the tent. I am surprised that the volume of the tent is that much bigger then the volume of this pyramid. It would help to draw the pyramid inside the tent to make a assessment whether our calculation is reasonable.
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There is a different way of looking at this tent. It turns out that the part of the tent which is in the vertical plane $y = 1-x$ is one-half of the ellipse with the long half-axes equal to $\sqrt{2}/2$ and the short half-axes equal to $1/2.$
To prove this claim verify the following identity $$ \frac{\left(x-\frac{1}{2}\right)^2 +\left(1-x-\frac{1}{2}\right)^2}{\left(\frac{\sqrt{2}}{2}\right)^2} + \frac{x(1-x)}{\left(\frac{1}{2}\right)^2} = 1 \quad \text{for all} \quad x \in [0,1]. $$
It turns out that the tent is formed by line segments which join the points on the above half-ellipse to the origin. Thus this tent is one-half of an elliptic cone, the volume of which equals ($(1/3) \times$ base $\times$ height) \[ \frac{1}{3} \, \left(\frac{1}{2} \, \frac{\sqrt{2}}{2} \, \frac{1}{2} \pi \right) \, \frac{\sqrt{2}}{2} = \frac{\pi}{24}. \] In the above formula we used the fact that the area of an ellipse with half-axis $a$ and $b$ is $ab\pi.$
Place the cursor over the image to start the animation.
- The content that we are reviewing is here.
- The information sheet
- We will start by reviewing Chapter 16. Do these problems
- The assigned problems for 16.1 are 1, 3, 7, 8-14, 15.
- The assigned problems for 16.2 are 1-33 (do most), 35, 36, 37, 38, 43, 44, 45, 51, 54, 55, 57, 61, 71, 72, 74, 75.
- The assigned problems for 16.3 are 1-8, 11, 12, 14, 15, 16-29, 32, 33, 34, 35, 36, 38, 44, 45, 46, 49, 52, 56, 57, 65-67, 68-71.
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I hope that you will enjoy the presentation of a visual proof that
\[
\int_0^1 x^2 dx = \frac{1}{3}
\]
which you can find in the pdf file posted in this item.
Right-click on the link in the next sentence and choose the option "Save as"; pay attention where you saved the file named Area_under_the_square_function.pdf. This is the pdf file that I demonstrated in class today. When you download the file Area_under_the_square_function.pdf to your computer, then open it in Acrobat reader in full screen mode. Browse through the file slowly by using "Page Down". To activate the animations you will have to pay attention to the content of the pages. The first animation appears on the 9th slide. You will recognize it by the text $y=x^2,$ $0\leq x\leq 1.$ The next animation appears on the 17th slide which contains Step 4. The next animation appears on the 24th and the 27th slide which both contain the text "... and set them in motion...". To activate an animation mouse left-click on the image.
This blueprint might be handy if you want to make yourself three identical pyramids which form the unit cube. Putting these three pyramids together to form the unit cube is a powerful demonstration that the volume of each one of them is $1/3$.