\documentclass[12pt]{article} %\pagestyle{empty} \setlength{\topmargin}{-1in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \usepackage{amssymb} \usepackage{amsmath} \usepackage{enumerate} \newcommand{\adj}{\mathrm{Adj}} \newcommand{\mbf}{\mathbf} \newcommand{\R}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\answer}{\textbf{Answer}} \newcommand{\ol}{\overline} \begin{document} {\Large Key: Take home 3} \begin{enumerate}[1.] \item Recall that the average value of a function of one variable on the interval $[a,b]$ is defined by \[ f_{ave} = \frac{\int_a^b f(x) \; dx}{b-a} \] Similarly, we can define average value for functions of two variables on a bounded set $D$. If $f(x,y)$ is a function of two variables, we define \[ f_{ave} = \frac{\iint_D f(x,y) \; dA}{\iint_D dA} \] \begin{enumerate}[(a)] \item Find the average value of $f(x,y)=e^{x+y}$ over the triangle with vertices $(0,0), (0,1), (1,0)$. \answer: Notice that the integral in the denominator is the area of the region, which for us is $\frac{1}{2}$, so there is no need to compute this integral. Next the only issue is really to find the limit of the integral. \begin{eqnarray*} f_{ave} & = & \frac{\iint_D f(x,y) \; dA}{1/2} = 2 \iint_D e^{x+y} \; dA \\ & = & 2\int_0^1 \int_0^{-x+1} e^{x+y} \; dy \; dx \\ & = & 2\int_0^1 e-e^x \; dx \\ & = & 2(1) = 2 \end{eqnarray*} \item How would you define average value for a function of three variables? \answer: \[ f_{ave} = \frac{\iiint_D f(x,y,z) \; dV}{\iiint_D dV} \] \item Find the average value of $f(x,y,z)=2x+3y+z$ over the domain $D$ defined by \[ 1 \leq x \leq 2 \qquad -2 \leq y \leq 1 \qquad 0 \leq z \leq 1 \] \answer: Again, the denominator is the volume of the domain, which for us is $3$. Thus, \begin{eqnarray*} f_{ave} & = & \frac{\iiint_D f(x,y,z) \; dV}{3} \\ & = & \frac{1}{3} \int_1^2 \int_{-2}^1 \int_0^1 (2x+3y+z) \; dz \; dy \; dx \\ & = & \frac{1}{3}(6) = 2 \end{eqnarray*} \end{enumerate} \item Consider the solid cylinder $x^2+y^2 \leq 1$, $1 \leq z \leq 2$ with density $\delta = (x^2+y^2)z^2$. \begin{enumerate}[(a)] \item Find the mass of the solid. \answer: Let $C$ be the solid cylinder. We need to compute \[ M = \iiint_C \delta \; dV \] We convert to cylindrical coordinate and we have $\delta = r^2z^2$. Our limits of integration will be: \[ 0 \leq \theta \leq 2\pi \qquad 0 \leq r \leq 1 \qquad 1 \leq z \leq 2 \] So we get: \begin{eqnarray*} M & = & \iiint_C \delta \; dV \\ & = & \int_1^2 \int_0^{2\pi} \int_0^1 r^2 z^2 r \; dr \; d\theta \; dz \\ & = & \frac{7\pi}{6} \end{eqnarray*} \item Find the center of mass of the solid. \answer: we need to find the integrals: \[ \ol{x}=\frac{\iiint_C x \delta \; dV}{M} \qquad \ol{y}=\frac{\iiint_C y \delta \; dV}{M} \qquad \ol{z}=\frac{\iiint_C z \delta \; dV}{M} \qquad \] We compute these using cylindrical coordinates. We first do $\ol{x}$: \begin{eqnarray*} \ol{x} & = & \frac{\iiint_C x \delta \; dV}{M} \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (r \cos \theta) (r^2 z^2) r \; dr \; d\theta \; dz \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^4 z^2 \cos \theta \; dr \; d\theta \; dz \\ & = & 0 \end{eqnarray*} (We could have seen this by using symmetry). Next for $\ol{y}$: \begin{eqnarray*} \ol{y} & = & \frac{\iiint_C y \delta \; dV}{M} \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (r \sin \theta) (r^2 z^2) r \; dr \; d\theta \; dz \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^4 z^2 \sin \theta \; dr \; d\theta \; dz \\ & = & 0 \end{eqnarray*} (We could have also seen this by using symmetry). Now, for $\ol{z}$: \begin{eqnarray*} \ol{z} & = & \frac{\iiint_C z \delta \; dV}{M} \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (z) (r^2 z^2) r \; dr \; d\theta \; dz \\ & = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^3 z^3 \; dr \; d\theta \; dz \\ & = & \frac{6}{7\pi} \left( \frac{15\pi}{8} \right) \\ & = & \frac{45}{28} \end{eqnarray*} So the center of mass is \[ \left( 0, 0, \frac{45}{28} \right) \] \end{enumerate} \item Let $E$ be the ellipsoid \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \] where $a,b,c$ are all positive. \begin{enumerate}[(a)] \item Find the volume of $E$. \answer: We need to compute: \[ \iiint_E dV \] To do this we first change variables: \[ x=au \qquad y=bv \qquad z=cw \] The Jacobian of this transformation is \begin{eqnarray*} J & = & \det \left[ \begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array} \right] = abc \end{eqnarray*} Our integral now becomes: \[ \iiint_E dV = \iiint abc \; du \; dv \; dw \] Notice that the transformation changes the ellipsoid into a sphere of radius 1. So, our integral (in $u,v,w$) is now over the unit ball $B$: $u^2+v^2+w^2 \leq 1$. So we now have \[ \iiint_E dV = \iiint_B abc \; du \; dv \; dw \] Now, this is perfect to convert it into spherical coordinates: \[ u= \rho \sin \phi \cos \theta \qquad v = \rho \sin \phi \sin \theta \qquad w = \rho \cos \phi \] And, of course our limits of integration will be \[ 0 \leq \rho \leq 1 \qquad 0 \leq \phi \leq \pi \qquad 0 \leq \theta \leq 2\pi \] So, finally we have: \begin{eqnarray*} \iiint_E dV & = & \int_0^\pi \int_0^1 \int_0^{2\pi} \rho^2 \sin \phi (abc) \; d\theta \; d\rho \; d\phi \\ & = & \frac{4\pi abc}{3} \end{eqnarray*} A short cut would have been to not convert into spherical coordinates, but to notice that $\iiint_B dV$ is the volume of the sphere which we know to be $4\pi/3$. \item Find the integral \[ \iiint_E \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \right) \; dV \] \answer: We make the same change of coordinates above (using the same Jacobian) and get: \begin{eqnarray*} \iint_B (u^2+v^2+w^2) (abc) \; du \; dv \; dw \end{eqnarray*} We now convert to spherical coordinates, as above and get \begin{eqnarray*} \int_0^\pi \int_0^1 \int_0^{2\pi} (\rho^2) \rho^2 \sin \phi (abc) \; d\theta \; d\rho \; d\phi & = & abc \int_0^\pi \int_0^1 \int_0^{2\pi} \rho^4 \sin \phi \; d\theta \; d\rho \; d\phi \\ & = & \frac{4 \pi a b c}{5} \end{eqnarray*} \end{enumerate} \item The cylinder $x^2+y^2=x$ divides the unit sphere ($x^2+y^2+z^2=1$) into two regions $S_1$ and $S_2$. $S_1$ is the region of the sphere inside the cylinder and $S_2$ is the region of the sphere outside the cylinder. \begin{enumerate}[(a)] \item Find the area of $S_1$. \answer: The upper surface of the sphere is given by the function \[ f(x,y) = \sqrt{1-x^2-y^2} \] We compute the partials \[ f_x = \frac{-x}{\sqrt{1-x^2-y^2}} \qquad f_y = \frac{-y}{\sqrt{1-x^2-y^2}} \] Then, the area of $S_1$ is given by \[ \frac{1}{2}A(S_1) = \iint_D \sqrt{1+f_x^2+f_y^2} \; dA \] Where $D$ is the disc in the $x-y$ plane bounded by the circle $x^2+y^2=x$. The formula is $\frac{1}{2}A(S_1)$ because $S_1$ has a top bottom halves and the integral is only the top half. The area of the bottom half is the same as the area of the top half. So, we substitute $f_x$ and $f_y$ into the integral and simplify: \[ \frac{1}{2}A(S_1) = \iint_{D} \frac{1}{\sqrt{1-x^2-y^2}} \; dA \] We now convert to polar coordinate. Notice that the equation $x^2+y^2=x$ transforms into the equation $r=\cos \theta$. Making $\theta$ go from $0$ to $\pi$ gives the whole circle and making $\theta$ go from $0$ to $\pi/2$ gives half the circle (graph it!). Next, using symmetry and geometry, notice that we can actually compute the integral only over half of the disc, the half with $y>0$ (and now we will have $\frac{1}{4}A(S_1)$ because we cut the area in half again). The reason I do this is because it made it easier for me to see the limits for integration in polar coordinate. \begin{eqnarray*} \frac{1}{4}A(S_1) & = & \iint_{D} \frac{1}{\sqrt{1-x^2-y^2}} \; dA \\ & = & \iint_{D} \frac{1}{\sqrt{1-r^2}} r \; dr \; d\theta \\ & = & \int_0^{\pi/2} \int_0^{\cos \theta} \frac{1}{\sqrt{1-r^2}} r \; dr \; d\theta \\ & = & \frac{\pi}{2} - 1 \end{eqnarray*} This gives \[ A(S_1) = 2\pi - 4 \] \item Find the area of $S_2$. \answer: we know the surface area of a sphere is $4\pi r^2$. So the surface area of the unit sphere is $4\pi$. Therefore the area of $S_2$ is \begin{eqnarray*} A(S_2) & = & A(\mbox{sphere}) - A(S_1) \\ & = & 4\pi - (2\pi-4) \\ & = & 2\pi +4 \end{eqnarray*} \end{enumerate} \end{enumerate} \end{document}