Math 234 — Final Exam Review

About double & triple integrals

The region for $I_1$ is the region that is between the circles with radius $a$ and$b$. In polar coordinates it is given by \[ 0\leq \theta \leq 2\pi, \quad a\leq r\leq b \] Using $x=r\cos \theta$, $y=r\sin\theta$, we get \begin{align*} I_1 &= \int_0^{2\pi} \int_{r=a}^b (r\cos \theta)^2\, r dr d\theta \\ &= \int_0^{2\pi} \int_{r=a}^b \cos^2 \theta \, r^3 dr d\theta \\ &= \int_0^{2\pi} \cos^2 \theta \bigl[\tfrac14r^4\bigr]_a^b d\theta \\ &= \frac14 \bigl(b^4-a^4\bigr)\int_0^{2\pi} \cos^2 \theta d\theta \\ &= \frac14 \bigl(b^4-a^4\bigr) \pi. \end{align*} (use the double angle formula for the last integral).

The region for $I_2$ is contained in the region for $I_1$. It consists of all points between the circles with radii $a$ and $b$ that lie above the graph of $y=c|x|$. This graph is v-shaped and consists of two half lines. The region is therefore given in poloar coordinates by \[ a\leq r\leq b, \quad \alpha\leq\theta\leq\beta, \] where the angles $\alpha$ and $\beta$ are determined by the slopes of the graph of $y=c|x|$. We have $\tan\alpha = c$, so $\alpha = \arctan c$; the other angle is $\beta = \pi-\alpha$. Thus we get \begin{align*} I_2 &= \int_\alpha^\beta \int_{r=a}^b (r\cos \theta)^2\, r dr d\theta \\ &= \int_\alpha^\beta \int_{r=a}^b \cos^2 \theta \, r^3 dr d\theta \\ &= \int_\alpha^\beta \cos^2 \theta \bigl[ \frac14r^4 \bigr]_a^b d\theta \\ &= \frac14 \bigl(b^4-a^4\bigr)\int_\alpha^\beta \cos^2 \theta d\theta \\ &= \frac14 \bigl(b^4-a^4\bigr) \int_\alpha^\beta \frac12 \{1+\cos 2\theta\} d\theta\\ &= \frac14 \bigl(b^4-a^4\bigr) \frac12 \{\beta-\alpha+\frac12\sin 2\beta-\frac12\sin2\alpha\} \end{align*} which one could simplify.

The region for $J_1$ is three dimensional. It is the region contained between the two spheres with radius $a$ and $b$. The best choice of coordinates is spherical coordinates. In these coordinates the region is described by \[ a\leq \rho\leq b,\quad 0\leq \phi\leq \pi, \quad 0\leq \theta\leq 2\pi. \] The coordinate $x$ is $x=\rho\sin\phi\cos\theta$, and the integral therefore is \begin{align*} J_1 &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi \int_{\rho=a}^b (\rho\sin\phi\cos \theta)^2\, \rho^2 \sin\phi d\rho d\phi d\theta \\ &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi \int_{\rho=a}^b \sin^3\phi\, \cos^2 \theta\, \rho^4 d\rho d\phi d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi}\int_{\phi=0}^\pi \sin^3\phi\, \cos^2 \theta\, d\phi d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta \left\{\int_{\phi=0}^\pi \sin^3\phi\, d\phi\right\} d\theta \end{align*} At this point we have to know these integrals, \[ \int_0^\pi \sin^3\phi\,d\phi \stackrel{u=-\cos \phi}= \int_{-1}^1 (1-u^2)du =\frac43 \] and \[ \int_0^{2\pi} \cos^2\theta\, d\theta = \pi. \qquad\text{(double angle trick)} \] We get \[ J_1 = \tfrac15\bigl(b^5-a^5\bigr) \frac43 \pi \]

The region for $J_2$ consists of all points between the spheres with radii $a$ and $b$ that also lie above the cone $z=\sqrt{x^2+y^2}$. The opening angle of the cone is $\pi/4 = 45^\circ$, so in spherical coordinates the region is described by \[ a\leq \rho \leq b, \qquad 0\leq \phi\leq \frac\pi4, \qquad 0\leq \theta \leq 2\pi. \] The integral is therefore almost the same as $J_1$ except one of the integration bounds are different: \begin{align*} J_2 &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4} \int_{\rho=a}^b (\rho\sin\phi\cos \theta)^2\, \rho^2 \sin\phi d\rho d\phi d\theta \\ &= \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4} \int_{\rho=a}^b \sin^3\phi\, \cos^2 \theta\, \rho^4 d\rho d\phi d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4} \sin^3\phi\, \cos^2 \theta\, d\phi d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta \left\{\int_{\phi=0}^{\pi/4} \sin^3\phi\, d\phi\right\} d\theta \end{align*} Again using $u=-\cos\phi$, we have \[ \int_0^{\pi/4} \sin^3\phi\,d\phi = \int_{-1}^{\sqrt2/2} (1-u^2)du =\left[u-\tfrac13u^3\right]_{-1}^{\sqrt2/2} =\tfrac5{12}\sqrt2 +\tfrac23 \] so we get \begin{align*} J_2 &= \tfrac15\bigl(b^5-a^5\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta \left\{\int_{\phi=0}^{\pi/4} \sin^3\phi\, d\phi\right\} d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr)\; \bigl(\tfrac5{12}\sqrt2 +\tfrac23\bigr) \int_{\theta=0}^{2\pi} \cos^2\theta d\theta \\ &= \tfrac15\bigl(b^5-a^5\bigr)\; \bigl(\tfrac5{12}\sqrt2 +\tfrac23\bigr) \pi. \end{align*} Switching the order of integration. The domain of integration in the integral \[ I = \int_0^{\infty} \int_0^x e^{-x}\;dy\, dx =\iint_\cR e^{-x}\;dA \] is given by \[ 0\leq y\leq x, \qquad 0\leq x\leq \infty \] These inequalities describe the region in terms of vertical slices: for each $x$ the $y$ values corresponding to points in $\cR$ satisfy $0 \leq y \leq x$. If you describe the region in terms of horizontal slices, then for any fixed $y$ we check which $x$ values correspond to points in $\cR$, and we find $y\leq x \le \infty$. So the other description of $\cR$ is \[ 0\le y\le \infty, \qquad y\leq x\le \infty. \] This gives us two ways of computing the integral. First, \begin{align*} I &= \int_0^{\infty} \int_0^x e^{-x}\;dy\, dx\\ &= \int_0^\infty e^{-x}\left\{ \int_0^xdy\right\} dx \quad\text{($e^{-x}$ is constant for the inner integral)}\\ &= \int_0^\infty xe^{-x}\;dx \quad\text{(integrate by parts)}\\ &= \left[(x-1)e^{-x}\right]_0^\infty\\ &=1. \end{align*} The other way is to first integrate with respect to $x$: \begin{align*} I &= \int_0^{\infty} \int_y^\infty e^{-x}\;dx\, dy\\ &= \int_0^{\infty} \left[-e^{-x}\right]_y^\infty dy\\ &= \int_0^{\infty} e^{-y}\;dy\\ &= 1. \end{align*}

Line integrals

Line integral $L_2$. The vector \[ \vAB = \begin{pmatrix} 4\\-3\end{pmatrix} \] is a tangent vector to the line segment $AB$, but it is not a unit vector. To get the unit tangent divide by the length of $AB$: \[ \vT=\frac{\vAB}{\|\vAB\|} = \begin{pmatrix} 4/5\\-3/5\end{pmatrix}. \] It follows that \[ \vF\cdot\vT = \frac 45 \cdot 2 +(-\frac35)(-1) = \frac{11}5, \] and the work integral \[ L_2 = \int_\cC \vF\cdot\vT \,ds = \frac{11}5\cdot\text{(length of $\cC$}) = 11. \]

Line integral $L_3$. To get the upward normal from the tangent, rotate it: \[ \vT = \begin{pmatrix} 4/5\\-3/5\end{pmatrix} \implies \vN = \begin{pmatrix} 3/5\\4/5\end{pmatrix} \] Thus \[ \vF\cdot\vN = (\frac 35)(2) +(\frac45)(-1) = \frac25, \] and \[ L_3 = \int_\cC \vF\cdot\vN\, ds = \frac25\,\bigl(\text{length $\cC$}\bigr) =\frac25\cdot 5 = 2. \]

Therefore, if we know that \[ \frac{\pd P}{\pd x} + \frac{\pd Q}{\pd y} = 0, \] then Green's theorem implies that \[ \int_\cC \vF\cdot\vN\, ds = 0. \] On the other hand, Green's theorem does not say anything about the work integral \[ \int_\cC \vF\cdot\vT\, ds. \]

For the problem in question, we have $A=(0,0)$, and $B=(\pi,1)$, and therefore \[ \int_\cC \vec\nabla f \cdot \vT\, ds = f(\pi,1) - f(0,0). \] The fact that the path $\cC$ is the straight line segment from $A$ to $B$ does not matter in this problem. The answer would have been the same for any other path.

The function is $f(x,y) = xy\cos(x)$, so \[ \int_\cC \vec\nabla f \cdot \vT\, ds = \pi\cdot 1\cdot \cos\pi - 0\cdot0\cdot\cos 0 = -\pi. \] Different solution: you could also calculate \[ \vec\nabla f = \begin{pmatrix} y\cos x-xy\sin y \\ x\cos x \end{pmatrix} \] parametrize the line segment $\cC$, and compute the (ugly) integral. This is in principle correct, and if you don't make mistakes, you should get the same answer. It was not the intended solution.