- Today in class I created this Mathematica file which should help with Problem 4 on the Mathematica Assignment. The file is called 20180823_MAP4.nb.
- In class I noticed that something was slightly off in the pictures that I was creating. It turns out that my solution of the equation \begin{equation*} 4 a E\Bigl(1 - \frac{b^2}{a^2} \Bigr) = 2 \pi \end{equation*} was wrong. Here $E(x)$ stands for the complete elliptic integral which in Mathematica is denoted by EllipticE[x]. The solution is clearly \begin{equation*} b = a \sqrt{1 - E^{-1}\Bigl(\frac{\pi}{2 a} \Bigr) }. \end{equation*} In class I overlooked $a$ in front of the square root. In fact, it is amazing that the pictures with the wrong formula for $b$ were so similar to the correct pictures.
- Please submit your Mathematica assignment over email by Monday noon. Before submitting your notebook save it with all output deleted. You can delete all output in a Mathematica notebook by clicking on "Delete All Output" in "Cell" menu item.
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When I assigned the first problem on the Mathematica assignment I did not realize that the beauty of the piece of the sphere that we study is emphasised when two or four of its copies are placed on the sphere. Recall that I asked you to explore the part of the sphere of radius 2 centered at the origin inside the cylinder of radius 1 with the axes $(1,0,z)$ with $z\in \mathbb R.$
Click on the image for several viewpoints of the object.
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In the picture below I added two cylinders in the half-sphere.
It is interesting to calculate the volume of the above solid. Recall that the sphere has the radius 2 and the orthogonal cylinders have radii equal to 1. It turns out that the volume is \begin{equation*} \frac{16}{9} (6 \pi -11) \approx 13.9548. \end{equation*}
- For Section 20.3 do Exercises 3, 6, 7, 8-11, 13-16, 23-26, 29, 30-35, 37, 38.
- We started Section 20.2 today. Do Exercises 3, 4, 5, 8, 9, 10, 12, 13, 14, 16, 20, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34.
- I updated my post of yesterday.
- We did Section 20.1 today. Do Exercises 5, 6, 9, 10, 11, 12, 16, 17, 19, 22, 24, 25-31.
- Today in class I created this Mathematica file which should help with Problem 1 on Mathematica assignment. The file is called 20180813_A1P1.nb.
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In this item I want to explain the connection between the geometric and the coordinate definition of the curl of a vector field.
Recall our discussion from August 1, 2018. We 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: $\vec{\nabla}$.
- Today we did Section 19.3. Do Exercises 2, 3, 4, 5, 6, 9, 11, 12, 13, 17, 18, 19, 20, 21, 24, 26, 27, 29, 34, 35, 36, 37, 38, 39, 40.
- Tomorrow we will do Section 19.4. Do Exercises 4, 5, 8, 12, 13, 14, 16, 17, 19, 21, 22, 24, 25, 27, 28, 30, 31, 32.
- Today we did Section 21.3. Do Exercises 1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19.
- Yesterday I created this Mathematica file illustrating Schoenhardt polyhedron, which is an important nonconvex polyhedron. The file is called Schoenhardt_polyhedron.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 CF 312.
- For Section 19.2 do Exercises 3, 4, 6, 7, 10, 12, 14, 15, 20, 22, 23, 26, 27, 29, 30, 33, 35, 38, 39, 42, 43, 45, 47, 49, 52, 56.
- For Section 19.1 do Exercises 7, 10, 11, 18, 20, 22, 25, 26, 34, 36, 38, 45, 46, 48, 50, 53, 55, 56, 57, 60, 61, 65.
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Today we talked about the concept of area vector. We stated a 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 class I presented a "flux" proof of the theorem for an arbitrary polyhedron.
- We did Problem 29 in Section 18.4 today. This problem deals with the Folium of Decartes curve. As pointed out in the book this curve is given by implicit equation \[ x^3+y^3 - 3 x y =0. \] The vector equation for this curve is \[ \left\langle \frac{3t}{1+t^3}, \frac{3t^2}{1+t^3} \right\rangle, \qquad t \in (-\infty,-1) \cup (-1,+\infty). \]
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I will explain one way to understand the graph of the Folium of Decartes without using the technology. 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 curve.
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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 to calculate the area in the first quadrant enclosed by this curve. 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 \vec{\jmath} \cdot d\vec{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 in improper integral and 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 in the picture above. This line integral equals \[ \int_0^{3/2} \left\langle 0, \frac{3}{2} - s \right\rangle \cdot \left\langle -1,-1 \right\rangle ds = \int_0^{3/2} \left( s - \frac{3}{2} \right) ds = -\frac{9}{8} \] Now we calculate the line integral along the dark blue part of the curve. The parameter $t \in [0,1]$ in this case. \begin{align*} \int_{C_b} x \vec{\jmath} \cdot d\vec{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 $2(15/8 - 9/8) = 3/2.$
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We also did Problem 25 in Section 18.4. 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*}
\vec{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 $\vec{F}$ is $\vec{0}.$ Since the vector field $\vec{F}$ is everywhere defined the curl test tells us that $\vec{F}$ is a gradient field. Therefore $\vec{F}$ 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.
Another way to calculate the requested line integral is to observe that \begin{equation*} \vec{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 $f(B) - f(A).$ - Today in class I created this Mathematica file. In this file I illustrated how to do the above two problems in Mathematica_8.
- From Review of Chapter 18 do Exercises 14, 21, 33, 39. 44, 45, 52, 54.
- Section 18.4 is a very interesting section. In class I tried to clarify the dependence of the Curl Tests for vector fields on the domain. For example, the vector field \[ \vec{F} = -\frac{y}{x^2+y^2} \vec{\imath} + \frac{x}{x^2+y^2} \vec{\jmath} \] is not a gradient field over its domain \[ D = \bigl\{(x,y) \in {\mathbb R}^2 : x^2 + y^2 \neq 0 \bigr\}. \] However, if we restrict the domain to \[ D^\prime = \bigl\{(x,y) \in {\mathbb R}^2 : y \neq 0 \ \text{or} \ x \gt 0 \bigr\}, \] then the vector field $\vec{F}$ is a gradient field. In fact, $\vec{F}$ 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 $D^\prime.$ Also, notice that $D^\prime$ is the plane from which the origin and the negative $x$-half-axis have been removed.
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The formula for $f(x,y)$ above looks complicated but this is not too complicated function. Here is its plot.
- Today in class I created this Mathematica file to illustrate that the vector field studied above is a gradient field.
- For Section 18.4 do Exercises 1, 2, 6, 7, 11, 14, 16, 20, 22, 24, 25 (the field involved in (b) is a gradient field), 26, 27, 28, 29, 30, 33, 34, 35, 37, 38.
- For Section 18.1 do Exercises 1-5, 10, 13, 14, 15, 17, 18, 20, 22, 25, 27, 29, 30, 32, 33, 37-41, 43, 45, 46, 47, 54-67.
- For Section 18.2 do Exercises 3, 5, 7, 13, 14, 16, 22, 23, 26-31, 32, 35, 40, 42-56.
- For Section 18.3 do Exercises 1, 3, 4, 13, 14, 16, 23, 25, 26, 28, 29, 31-34, 36, 39, 40, 42, 44, 53.
- We started Section 18.1. Do Exercises 1-5, 10, 13, 14, 15, 17, 18, 20, 22, 25, 27, 29, 30, 32, 33, 37-41, 43, 45, 46, 47, 54-67.
- Today in class I created this Mathematica file to illustrate in Mathematica the change of variables problem that we did in class.
- For Section 21.2 do Exercises 5, 6, 9, 10, 11, 13, 14, 15, 16.
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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 $S_a.$
- Evaluate the double integral $\iint_R y e^{\sqrt{x}} dA$ where $R$ is the finite region in the first quadrant bounded by the graph of $y=\sqrt{x}$ and the verticla line $y=1.$
- For Section 17.3 do Exercises 1-10, 12, 15, 16, 17, 19, 20, 21, 27, 28, 29, 32, 33, 34, 36, 37, 38.
- For Section 17.4 do Exercises 6, 7, 8, 9, 16, 17, 19, 20, 21, 22, 23, 26, 29, 36, 37.
- For Review Chapter 17 do Exercises 12, 23, 25, 26, 36, 38, 39, 41, 42, 52.
- I created this Mathematica file to illustrate how Mathematica plots Vector Fields. The file is called VectorFields_v8.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 CF 312. Open Mathematica first, then open VectorFields_v8.nb from Mathematica. Here is a pdf printout of the above Mathematica file.
- Today in class I created this Mathematica file. The file is called 20180712_Surfaces.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.
- Today we did Section 21.1 do Exercises 2,5-12, 14-22, 27-30, 32-35, 37-48
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Today in class I created
this Mathematica file. The file is called 20180711_Curves.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. Look in
Start -> All Programs -> Wolfram Mathematica -> Mathematica 8. Open Mathematica first, then open 20180711_Curves.nb from Mathematica. - More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.
- We finished Section 17.2. Do Exercises 5, 6, 10, 12, 15, 16, 18, 19, 21, 22, 24, 27, 28, 30, 32, 40 (the direction of the motion of the wheel is wrong; it should be $\vec{\imath}$), 43
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Below I list a 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 \[ x(t) = (1+\cos t)\cos t, \quad y(t) = (1+\cos t) \sin t \quad \text{where} \quad 0 \leq t \leq 2\pi. \]
- In the problems above I mentioned three famous curves. Here is Wikipedia's List of curves. For more interesting links see External links at the end of that Wikipedia page.
- We started Section 17.2 today. Do Exercises 5, 6, 10, 12, 15, 16, 18, 19, 21, 22, 24, 27, 28, 30, 32, 40 (the direction of the motion of the wheel is wrong; it should be $\vec{\imath}$), 43
- We started Section 17.1 today. Do Exercises 2, 3, 5, 8, 30, 34, 37, 40, 49, 50, 51, 52, 54, 62, 68, 72, 82.
- On Thyrsday I handed out the first assignment which will be graded as the first part of Exam 1. On Thursday I emalied a revised version of this assignment. Two questions on this assignment involve quite difficult integrals. The integral in Problem 2 is difficult, but Mathematica evaluates it easily. The integral in Problem 3(i) is also difficult. One can set up this integral in two different ways. One of these integrals evan Mathematica can not evaluate exactly.
-
On Tuesday I mentioned the following question: What is the average distance of a point in a unit disk to a fixed point on its perimeter? Let $D$ be the unit disk centered at the origin and let $(1,0)$ be the fixed point on its perimeter. Becase of the symmetry the average distance does not depend on the special choice of this point. The average distance is
$$
\frac{1}{\pi} \iint_D \sqrt{(x-1)^2+y^2}dA = \frac{1}{\pi} \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{(x-1)^2+y^2}dy \, dx .
$$
The double integral involved is the volume below.
Integrate[
Integrate[
((x-1)^2+y^2)^(1/2), {y,-(1-x^2)^(1/2),(1-x^2)^(1/2)},
Assumptions->And[x>-1,x<1]
],
{x,-1,1}
]
- Today we discussed the concept of the average value of a function. Below we present several problems involving average distance. These problems involve double and triple integrals. You can state and solve analogous problems involving single integrals.
-
What is the average distance of a point in a unit disk to disk's center? Let $D$ be the unit disk centered at the origin. The average distance is $(1/\pi) \iint_D \sqrt{x^2+y^2}dA.$ The double integral involved is the volume below.
-
In the previous item we used so called Euclidean distance to calculate the distance from the origin. Another popular distance, which is more alike what we experience in everyday life, is so called taxicab distance. The taxicab distance of the point $(x,y)$ to the origin is $|x|+|y|.$ Thus, the average taxicab distance of a point in a unit disk $D$ to the center of the unit disk is
$$
\frac{1}{\pi} \iint_D (|x|+|y|)dA
$$
The preceding double integral is the volume below.
-
The calculation in the previous item becomes simpler if we replace the unit disk with the square with the corners: $(-1,-1)$, $(-1,1)$, $(1,1)$, $(-1,1).$ Call this square $S.$ Using the taxicab distance find the average distance of the point $(x,y)$ in the square $S$ to the origin. The average distance of a point in the square $S$ to the origin is
$$
\frac{1}{4} \iint_S (|x|+|y|)dA
$$
The double integral involved is the volume below.
-
The calculation becomes more involved if we replace the taxicab distance with the Euclidean distance in the square $S$ with the corners: $(-1,-1),$ $(-1,1),$ $(1,1),$ $(-1,1).$ Using the Euclidean distance the average distance of a point in the square $S$ to the origin is
$$
\frac{1}{4} \iint_S \sqrt{x^2 + y^2}\, dA
$$
The double integral involved is the volume below.
- Next, it would be interesting to repeat all these calculations by replacing the unit disk with the unit ball and the square $S$ with the cube $C$ with the corners at the coordinates at the eight points involving $1$ and $-1.$ In the items below we will use $B$ to denote the unit ball. Recall that the volume of the unit ball is $4\pi/3.$
- Find the average Euclidean distance of a point in the unit ball $B$ to the origin. This average is given by the integral $$ \frac{3}{4\pi} \iiint_B \sqrt{x^2 + y^2+z^2}\, dV . $$ The triple integral is calculated in spherical coordinates as \begin{align*} \iiint_B \sqrt{x^2 + y^2+z^2}\, dV & = \int_0^1 \int_0^{2\pi} \int_0^{\pi} \rho^3 \sin \phi \, d\phi d\theta d\rho \\ & = 2\pi \ \frac{1}{4} (\rho^4) \Bigl. \Bigr|_0^1 \ ( -\cos \phi)\Bigl. \Bigr|_0^1 \\ & = \pi \end{align*} Thus the average is $3/4.$
- Find the average taxicab distance of a point in the unit ball $B$ to the origin. This average is given by the integral $$ \frac{3}{4\pi} \iiint_B (|x|+|y|+|z|)\, dV . $$ Denote by $B_1$ the part of the unit ball which is in the first octant. The triple integral is calculated in spherical coordinates as \begin{align*} \iiint_B (|x|+|y|+|z|)\, dV & = 8 \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 ((\cos \theta) (\sin \phi)+(\sin \theta) (\sin \phi)+ \cos\phi ) \rho^3 \sin \phi \, d\rho d\phi d\theta \\ & = 2 (\rho^4) \Bigl. \Bigr|_0^1 \int_0^{\pi/2} \int_0^{\pi/2} \left( (\cos \theta) (\sin \phi)^2+(\sin \theta) (\sin \phi)^2+ (\cos\phi) (\sin \phi) \right) d\phi d\theta \\ & = 2 \left( \frac{\pi}{4} + \frac{\pi}{4} + \frac{\pi}{4} \right) \\ &= \frac{3\pi}{2} \end{align*} Thus the average is $9/8.$
- Find the average taxicab distance of a point in the cube $C$ to the origin. This average is given by the integral $$ \frac{1}{8} \iiint_C (|x|+|y|+|z|)\, dV . $$ Denote by $C_1$ the part of the cube $C$ which is in the first octant. The triple integral is calculated in Cartesian coordinates \begin{align*} \iiint_C (|x|+|y|+|z|)\, dV & = 8 \int_0^{1} \int_0^{1} \int_0^1 (x+y+z ) \, dx dy dz \\ & = 8 \left( \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \right) \\ &= 12 \end{align*} Thus the average is $3/2.$
- Finally, let us find the average Euclidean distance of a point in the cube $C$ to the origin. This average is given by the expression $$ \frac{1}{8} \iiint_C \sqrt{x^2+y^2+z^2}\, dV . $$ Denote by $P$ the following pyramid: $$ P = \bigl\{ (x,y,z) \in {\mathbb R}^3 \, | \, 0\leq x \leq 1, 0 \leq y \leq x, 0 \leq z \leq x\bigr\}. $$ The cube $C$ consists of $24$ pyramids that are congruent to the pyramid $P.$ We will calculate the triple integral in spherical coordinates. For that we need to describe $P$ in spherical coordinates: $$ P = \left\{ (\rho,\theta,\phi) \, | \, 0 \leq \theta \leq \frac{\pi}{4}, \arctan\left(\frac{1}{\cos \theta}\right) \leq \phi \leq \frac{\pi}{2}, 0 \leq \rho \leq \frac{1}{(\cos \theta)(\sin \phi)}\right\}. $$ Now the integral is calculated as follows: \begin{align*} \iiint_C \sqrt{x^2+y^2+z^2}\, dV & = 24 \iiint_P \sqrt{x^2+y^2+z^2}\, dV \\ & = 24 \int_0^{\pi/4} \int_{\arctan\bigl(1/(\cos\theta)\bigr)}^{\pi/2} \int_0^{1/((\cos\theta)(\sin\phi))} \rho^3 (\sin \phi) d\rho d\phi d\theta \\ &= 6 \int_0^{\pi/4} \int_{\arctan\bigl(1/(\cos\theta)\bigr)}^{\pi/2} \frac{1}{ (\cos\theta)^4(\sin\phi)^3}d\phi d\theta \\ &= 6 \int_0^{\pi/4} \frac{1}{ (\cos\theta)^4} \int_{\arctan\bigl(1/(\cos\theta)\bigr)}^{\pi/2} \frac{\sin \phi}{ (\sin\phi)^4}d\phi d\theta \\ &= 6 \int_0^{\pi/4} \frac{1}{ (\cos\theta)^4} \int^{\arctan\bigl(1/(\cos\theta)\bigr)}_{\pi/2} \frac{d(\cos \phi)}{ (1-(\cos\phi)^2)^2} \, d\theta \\ &= 6 \int_0^{\pi/4} \frac{1}{ (\cos\theta)^4} \int^{\frac{\cos\theta}{\sqrt{1+(\cos\theta)^2}}}_{0} \frac{du}{ (1-u^2)^2} \ d\theta \\ &= 6 \int_0^{\pi/4} \frac{1}{ (\cos\theta)^4} \left(\frac{u}{2(1-u^2)} +\frac{1}{4} \ln\left(\frac{1+u}{1-u}\right)\right) \Biggl.\Biggr|^{\frac{\cos\theta}{\sqrt{1+(\cos\theta)^2}}}_{0} \ d\theta \\ &= 3 \int_0^{\pi/4} \left( \frac{\sqrt{1+(\cos\theta)^2}}{(\cos\theta)^3} +\frac{1}{2} \frac{1}{(\cos\theta)^4} \ln \left(\frac{\sqrt{1+(\cos\theta)^2} +\cos\theta }{\sqrt{1+(\cos\theta)^2} -\cos\theta}\right)\right) \ d\theta \\ &= 3 \int_0^{\pi/4} \left( \frac{\sqrt{1+(\cos\theta)^2}}{(\cos\theta)^3} + \frac{1}{(\cos\theta)^4} \ln \left( \sqrt{1+(\cos\theta)^2} +\cos\theta \right)\right) \ d\theta \\ \end{align*} This is the best I could do by hand. These two integrals are difficult to do by hand. Mathematica gives $$ \int_0^{\pi/4} \frac{\sqrt{1+(\cos\theta)^2}}{(\cos\theta)^3} \ d\theta = \frac{1}{2}\left(\sqrt{3} + \ln\bigl(2+\sqrt{3}\bigr)\right) $$ and $$ \int_0^{\pi/4} \frac{1}{(\cos\theta)^4} \ln \left( \sqrt{1+(\cos\theta)^2} +\cos\theta \right) \ d\theta = -\frac{\pi}{9} +\frac{1}{6}\left(\sqrt{3} + \ln\bigl(362+209\sqrt{3}\bigr) \right) $$ Thus the volume is $$ -\frac{\pi}{3} + 2 \sqrt{3} +\frac{1}{2} \ln\bigl(18817+10864\sqrt{3}\bigr) \approx 7.68474 $$ and the average distance is $$ -\frac{\pi}{24} + \frac{\sqrt{3}}{4} +\frac{1}{16} \ln\bigl(18817+10864\sqrt{3}\bigr) \approx 0.960592 $$
- Today we reviewed integration in polar coordinates. The most important concept here is the area element in polar coordinates: $dA = r \, d\theta \, d r$.
- 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: $dV = \rho^2 (\sin \phi) \, d\theta\, d\phi \, d\rho$.
-
To understand the volume element in spherical coordinates one needs to understand the spherical coordinates first. The first step in understanding spherical coordinates is to understand the parametrization of the unit vectors in $\mathbb R^3$ using angles $\theta$ and $\phi$. I hope that the animation below will be helpfull:
-
The image below should help you to understand the element of volume in spherical coordinates formula $dV = \rho^2 (\sin \phi) \, d\theta\, d\phi \, d\rho$.
- Here is a pdf of Chapter 16.
- The information sheet
- We will start by reviewing Chapter 16. Do these problems
- The assigned problems for 16.1 are 1, 3, 6-11, 13.
- The assigned problems for 16.2 are 1-27 (do most), 32, 33, 34, 35, 41, 42, 43, 44, 53, 54, 56, 58, 59, 60.
- The assigned problems for 16.3 are 1-4, 5, 7, 8, 11, 12, 14-18, 19-27, 35, 37, 39, 43, 45, 49, 50, 51, 52, 60-62, 59, 63, 64
- The assigned problems for 16.4 are 1-8, 9, 12-15, 16-22, 25, 27, 28, 31, 33, 34, 35 (but do evaluate the area), 36
- The assigned problems for 16.5 are 1-20 (all, but in particular 9, 10, 11, 19, 20), 19-20 (and evaluate the integrals), 21-23, 24-28, 29-34, 36, 37, 38, 39, 42, 43, 48, 49, 52, 56
- Some relevant Chapter 16 Review exercises and problems are 17, 21, 23, 63, 64.
- As I indicated today we will be using Mathematica in this class. We will be using Mathematica 8 which is available in BH 215 and CF 312. To get started with Mathematica visit my Mathematica page. There are several links to Wolfram's movies that are an easy introduction to this software.