# Summer 2017 MATH 225: Multivariable Calculus and Geometry. II Branko Ćurgus

Monday, August 14, 2017

Tuesday, August 8, 2017

• For Section 20.2 do Exercises 3, 4, 5, 8, 9, 10, 12, 13, 14, 16, 20, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34.
• For Section 20.3 do Exercises 3, 6, 7, 8-11, 13-16, 23-26, 29, 30-35, 37, 38.

Monday, August 7, 2017

• For Section 20.1 do Exercises 5, 6, 9, 10, 11, 12, 16, 17, 19, 22, 24, 25-31.

Tuesday, August 1, 2017

• For the coming exam we covered the following topics.
• 21.1: Parametric equations of surfaces (planes, graphs, cylinders, spheres, surfaces of revolution)
• 21.2: General change of coordinates in double and triple integrals, the concept and the meaning of the Jacobian.
• 17.3, 17.4: The concept of a vector field and the concept of the flow of a vector field (in particular determining a flow line through a given point of a given vector field)
• Chapter 18: the concept of line integral (its physical meaning and how to evaluate it), the fundamental theorem of calculus for line integrals, the concept of the scalar curl, Green's theorem and the curl tests
• Chapter 19, 21.3: the concept of flux integral (how to estimate it and how to evaluate it exactly), the concept of divergence and the divergence theorem

Monday, July 31, 2017

• For 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.
• For Section 19.4 do Exercises 4, 5, 8, 12, 13, 14, 16, 17, 19, 21, 22, 24, 25, 27, 28, 30, 31, 32.
• We agreed to move the second exam to Thursday, August 3, 2017.

Thursday, July 27, 2017

• 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 21.3 do Exercises 1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19.

Wednesday, July 26, 2017

• 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.

Thursday, July 25, 2017

• 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 $\left\langle \frac{3t}{1+t^3}, \frac{3t^2}{1+t^3} \right\rangle, \qquad t \in (-\infty,-1) \cup (-1,+\infty).$
• 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.
• 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 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 asked to calculate the area in the first quadrant enclosed by the 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 that 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 $2(15/8 - 9/8) = 3/2.$
• From Review of Chapter 18 do Exercises 14, 21, 33, 39. 44, 45, 52, 54.

Friday, July 21, 2017

• 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.
• 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.
• The formula for $f(x,y)$ above looks complicated but this is not too complicated function. Here is its plot.

Thursday, July 20, 2017

• 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.

Tuesday, July 19, 2017

• 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.

Monday, July 17, 2017

• 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.
• After class I asked the following question: What is the average distance between two points chosen in the unit interval $[0,1].$ I created this Mathematica file to illustrate how Mathematica approximates this average.

Friday, July 14, 2017

• 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.

Thursday, July 13, 2017

• For Section 21.2 do Exercises 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 $S_a.$
• Yesterday we discussed the question few more questions involving average distance.
• For example: What is the average distance of a point in the unit disk to the union of the coordinate axes? Let $D$ be the unit disk centered at the origin. The distance of a point $(x,y)$ to the union of the coordinate axes is given by $\min\{|x|, |y|\}.$ Therefore, the average distance of a point in $D$ to the union of the coordinate axes is $\frac{1}{\pi} \iint_D \min\{|x|, |y|\} \, dA.$ The preceding double integral is the volume below.
To evaluate this volume we is use the symmetries to deduce that the volume equals in the polar coordinates $8 \int_0^{\pi/4} \int_0^1 r^2 \sin(\theta) \,dr d\theta = \frac{4}{3}\bigl(2-\sqrt{2}\bigr).$ Thus the average distance of a point in the unit disk to the union of the coordinate axes is $\frac{4}{3\pi}\bigl(2-\sqrt{2}\bigr) \approx 0.248615.$
• An easier example: What is the average distance of a point in the square with vertices $(1,1)$, $(-1,1)$, $(-1,-1),$ and $(1,-1)$ to the union of the coordinate axes? Let $S$ denote this square. The distance of a point $(x,y)$ to the union of the coordinate axes is given by $\min\{|x|, |y|\}.$ Therefore, the average distance of a point in $S$ to the union of the coordinate axes is $\frac{1}{4} \iint_S \min\{|x|, |y|\} \, dA.$ The preceding double integral is the volume below.
This volume consists of four pyramids. Thus the volume is $4 \times \frac{1}{3} \times 1 \times 1 \times 1 = \frac{4}{3}.$ Thus the average distance of a point in $S$ to the union of the coordinate axes is $\frac{1}{3}.$
• An example with a different square: What is the average distance of a point in the square with vertices $(1,0)$, $(0,1)$, $(-1,0),$ and $(0,-1)$ to the union of the coordinate axes? Let $S_0$ denote this square. The distance of a point $(x,y)$ to the union of the coordinate axes is given by $\min\{|x|, |y|\}.$ Therefore, the average distance of a point in $S_0$ to the union of the coordinate axes is $\frac{1}{2} \iint_{S_0} \min\{|x|, |y|\} \, dA.$ The preceding double integral is the volume below.
This volume consists of four pyramids. Thus the volume is $4 \frac{1}{3} \left(\frac{1}{2} \frac{1}{2} \sqrt{2} \right) \frac{\sqrt{2}}{2} = \frac{1}{3}.$ Thus the average distance of a point in $S$ to the union of the coordinate axes is $\frac{1}{6}.$
• It is important to notice that it is clear that the average calculated for the disk is between the averages calculated for two squares. In fact the average of two averages for the squares is very close to the average for the disk since $\frac{ \frac{1}{3}+ \frac{1}{6}}{2} = \frac{1}{4}.$

Wednesday, July 12, 2017

• Today we did Section 21.1 do Exercises 2,5-12, 14-22, 27-30, 32-35, 37-48

Friday, July 7, 2017

• For 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
• I updated the Mathematica file that I created yesterday. Now it includes animations that illustrate vector valued functions as a point in motion.
• 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 $x(t) = (1+\cos t)\cos t, \quad y(t) = (1+\cos t) \sin t \quad \text{where} \quad 0 \leq t \leq 2\pi.$

Thursday, July 6, 2017

• We did Section 17.1 today. Do Exercises 2, 3, 5, 8, 30, 34, 37, 40, 49, 50, 51, 52, 54, 62, 68, 72, 82.
• Today in class I created this Mathematica file. The file is called 20170706.nb for today's date written as an eight-digit integer.

Wednesday, July 5, 2017

• Today in class we solved Problem 61 in Section 16.3. The region of integration $W$ in this problem is given in ${\mathbb R}^3.$ In the set notation this region of integration can be written as $W = \Bigl\{ (x,y,z) \in {\mathbb R}^3 \, |\, 0 \leq x \ \ \text{and} \ \ 0 \leq y \ \ \text{and} \ \ 0 \leq z \ \ \text{and} \ \ x^2 + 3 y^2 \leq 12 \ \ \text{and} \ \ z \leq 6 y^2 \Bigr\}$ The conditions $0 \leq x \ \ \text{and} \ \ 0 \leq y \ \ \text{and} \ \ 0 \leq z$ tell us that the point $(x,y,z)$ is in the first octant. The condition $x^2 + 3 y^2 \leq 12$ tell us that the point $(x,y,z)$ is in the elliptic cylinder $x^2 + 3 y^2 = 12$. The condition $z \leq 6y^2$ tell us that the point $(x,y,z)$ is bellow the parabolic cylinder $z= 6 y^2.$ The easiest way to adopt the above set description of $W$ to a description suitable for triple integration is to fix $y$ such that $0 \leq y \leq 2$ and read the condition for $x$ from the equation of the ellipse: $W = \Bigl\{ (x,y,z) \in {\mathbb R}^3 \, |\, 0 \leq y \leq 2 \ \ \text{and} \ \ 0 \leq z \leq 6y^2 \ \ \text{and} \ \ 0 \leq x \leq \sqrt{ 12 - 3 y^2} \Bigr\}.$ Mathematica calculates that the volume of this region in ${\mathbb R}^3$ is $\int_0^2 \int_0^{ 6y^2} \int_0^{\sqrt{12-3y^2}} 1 \ dx\, dz \, dy = 6\sqrt{3} \, \pi.$ The only difficult part in calculation of this integral is the following antiderivative: $\int y^2 \sqrt{4-y^2} dy = \frac{1}{4} y (y^2-2) \sqrt{4-y^2} + 2 \arcsin(y/2) + C.$
• Today in class I created this Mathematica file. The file is called 20170705.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 20170705.nb from Mathematica.
• More information on how to use Mathematica version 8 you can find on my Mathematica 8 page.

Friday, June 30, 2017

• Today we discussed the question of average value of a function of two and three variables. Basic questions related to this concept are questions involving average distance.
• For example: What is the average distance of a point in a unit disk to its 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.
To evaluate this volume we do not need to do iterated integrals since we know that the volume of the cone that is being removed is $\pi/3$. Thus the pictured volume is $2\pi/3$. Thus, the average distance is $2/3$.
• 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.
To calculate this double integral we use the polar coordinates. To simplify the taxicab distance we consider only one quarter of the unit disk $D_1$ which is in the first quadrant: $$\iint_D (|x|+|y|)dA = 4 \iint_{D_1} (x+y)dA = 4 \int_0^{\pi/2} \int_0^1 \bigl( \cos \theta + \sin \theta \bigr) r^2 dr d\theta = \frac{8}{3}.$$ Thus, the average value is $8/(3\pi) \approx 0.848826.$
• 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.
Now notice the symmetry of the part of this volume in the first octant. The part above the level $z=1$ is a pyramid which is congruent to the pyramid that is missing below the level $z=1$. Thus the pictured volume is $4$. So, the average distance is exactly $1$.
• 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.
Before doing iterated integrals recall that $$\int \frac{1}{(1 - u^2)^2}\, du = \frac{u}{2(1-u^2)} +\frac{1}{4} \ln\left(\frac{1+u}{1-u}\right)$$ The calculation of the double integral here is a little tricky. Denote by $S_0$ the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$. The description of this triangle in polar coordinates is $$S_0 = \Bigl\{ (r, \theta) \, | \, 0\leq \theta \leq \pi/4, \ 0\leq r \leq \frac{1}{\cos \theta} \Bigr\}$$ Notice that the square $S$ consists of $8$ triangles which are congruent to $S_0$. The volume pictured above has several symmetries which allow us to calculate in polar coordinates: \begin{align*} \iint_S \sqrt{x^2 + y^2}\, dA & = 8 \iint_{S_0} \sqrt{x^2 + y^2}\, dA \\ & = 8 \int_0^{\pi/4} \int_0^{1/(\cos \theta)} r^2 dr \, d\theta \\ & = \frac{8}{3} \int_0^{\pi/4} \frac{1}{(\cos \theta)^3}\, d\theta \\ & = \frac{8}{3} \int_0^{\pi/4} \frac{\cos \theta}{(\cos \theta)^4}\, d\theta \\ & = \frac{8}{3} \int_0^{\pi/4} \frac{1}{(1 - (\sin \theta)^2)^2}\, d(\sin \theta) \\ & = \frac{8}{3} \int_0^{1/\sqrt{2}} \frac{1}{(1 - u^2)^2}\, du \\ & = \frac{2}{3} \bigl(2\sqrt{2} + \ln\bigl(3+2\sqrt{2}\bigr) \bigr) \approx 3.06078 \end{align*} Thus, the average of the the Euclidean distance of a point in the square $S$ to the origin is $$\frac{1}{6} \bigl(2\sqrt{2} + \ln\bigl(3+2\sqrt{2}\bigr) \bigr) \approx 0.765196$$
• 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 $$-\frac{\pi}{24} + \frac{\sqrt{3}}{4} +\frac{1}{16} \ln\bigl(18817+10864\sqrt{3}\bigr) \approx 0.960592$$

Wednesday, June 28, 2017

• Yesterday and 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 $dV = \rho^2 (\sin \phi) \, d\theta\, d\phi \, d\rho$.

Monday, June 26, 2017

• 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 we 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.