Homework 12

1. (50 points)
   Compute the following integrals using polar coordinates. In each case sketch the region of integration.
   (a) \(\int_S (x+y) \, dA\), where \(S = \{(x,y) \in \mathbb{R}^2 : 1 \le x^2 + y^2 \le 4, x \le 0\}\).

   ​

   Let \(g:(r, \theta) \mapsto (x,y) = (r\cos\theta, r\sin\theta)\). Then \(\iint_S f dA = \iint_D f(g(r,\theta)) |J_g(r,\theta)| dr d\theta\)
   ​\(= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_1^2 (r\cos\theta + r\sin\theta) \cdot r dr d\theta = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left( \frac{1}{3} r^3 \cos\theta + \frac{1}{3} r^3 \sin\theta \right) \Big|_1^2 d\theta\)
   ​\(= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{7}{3} \cos\theta + \frac{7}{3} \sin\theta d\theta = \frac{7}{3} (\sin\theta - \cos\theta) \Big|_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \frac{7}{3} (-1 - 1) = -\frac{14}{3}\)
   (b) \(\int_S \arctan\left(\frac{y}{x}\right) \, dA\), where \(S = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \ge 1, x^2 + y^2 \le 4, 0 \le y \le x\}\).

   ​

   Let \(g:(r, \theta) \mapsto (x, y) = (r \cos \theta, r \sin \theta)\). Then \(\iint_S f \, dA = \iint_D f(g(r, \theta)) |J_g(r, \theta)| \, dr \, d\theta\)
   ​\(= \int_0^{\pi/4} \int_1^2 \arctan\left(\frac{\sin \theta}{\cos \theta}\right) \cdot r \, dr \, d\theta\) \(= \int_0^{\pi/4} \int_1^2 \theta \cdot r \, dr \, d\theta\) \(= \int_0^{\pi/4} \left[ \frac{\theta}{2} r^2 \right]_1^2 \, d\theta\)
   ​\(= \int_0^{\pi/4} \left( \frac{\theta}{2} \cdot 4 - \frac{\theta}{2} \cdot 1 \right) \, d\theta\) \(= \int_0^{\pi/4} \frac{3}{2} \theta \, d\theta\) \(= \frac{3}{4} \theta^2 \Big|_0^{\pi/4}\) \(= \frac{3}{4} \times \frac{\pi^2}{16}\) \(= \frac{3\pi^2}{64}\)
   (c) \(\int_S y(x^2+y^2)^{-\frac{5}{2}} \, dA\), where \(S\) is the planar region lying above the \(x\)-axis, bounded on the left by \(x=1\) and above by \(x^2+y^2=2\).

   ​

   Let \(g: (r,\theta) \mapsto (x,y) = (r \cos\theta, r \sin\theta)\). Then \(\iint_S f \,dA = \iint_D f(g(r,\theta)) |J_g(r,\theta)| \,dr \,d\theta\)
   Since \(0 \le \theta \le \frac{\pi}{4}\), \(\frac{1}{\cos\theta} \le r \le \sqrt{2}\), then \(\iint_S f \,dA = \int_0^{\frac{\pi}{4}} \int_{\frac{1}{\cos\theta}}^{\sqrt{2}} r \sin\theta (r^2)^{-\frac{5}{2}} \cdot r \,dr \,d\theta\)
   ​\(= \int_0^{\frac{\pi}{4}} \int_{\frac{1}{\cos\theta}}^{\sqrt{2}} r^3 \sin\theta \,dr \,d\theta = \int_0^{\frac{\pi}{4}} \left( -\frac{1}{2} r^{-2} \sin\theta \right) \Big|_{\frac{1}{\cos\theta}}^{\sqrt{2}} \,d\theta = \int_0^{\frac{\pi}{4}} \left( -\frac{1}{4} + \frac{1}{2} \cos^2\theta \right) \sin\theta \,d\theta\)
   ​\(= \int_0^{\frac{\pi}{4}} -\frac{1}{4} \sin\theta \,d\theta + \frac{1}{2} \int_0^{\frac{\pi}{4}} \cos^2\theta \sin\theta \,d\theta = \frac{1}{4} \cos\theta \Big|_0^{\frac{\pi}{4}} + \frac{1}{2} \int_1^{\frac{\sqrt{2}}{2}} t^2 (-dt) = \frac{\sqrt{2}}{8} - \frac{1}{4} + \frac{1}{6} - \frac{\sqrt{2}}{24}\)
   ​\(= \frac{3\sqrt{2} - 6 + 4 - \sqrt{2}}{24} = \frac{2\sqrt{2} - 2}{24} = \frac{\sqrt{2} - 1}{12}\)
   (d) \(\int_0^1 \int_y^1 \frac{xe^x}{x^2+y^2} \, dxdy\).

   ​

   Let \(g: (r, \theta) \mapsto (x, y) = (r\cos\theta, r\sin\theta)\), then \(\int_{0}^{1} \int_{y}^{1} \frac{x e^{x}}{x^{2}+y^{2}}dx dy = \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{1}{\cos\theta}}\frac{r\cos\theta \cdot e^{r\cos\theta}}{r^{2}}\cdot r dr d\theta\)

   \(= \int_0^{\frac{\pi}{4}} \int_0^{\frac{1}{\cos\theta}} \cos\theta \cdot e^{r\cos\theta} dr d\theta = \int_0^{\frac{\pi}{4}} e^{r\cos\theta} \bigg|_0^{\frac{1}{\cos\theta}} d\theta = \int_0^{\frac{\pi}{4}} (e-1)d\theta\)

   \(= (e-1)\theta \bigg|_0^{\frac{\pi}{4}} = (e-1)\frac{\pi}{4}\)
   (e) \(\int_0^2 \int_0^{\sqrt{2x-x^2}} xy \, dydx\)

   ​

   \(\int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} xy \,dy\,dx = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\sin 2\theta}{\sin \theta}} r^3 \sin \theta \cos \theta \,dr\,d\theta\) \(= \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta \cos \theta}{4} \cdot r^4 \Big|_{0}^{\frac{\sin 2\theta}{\sin \theta}} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2\theta}{8} \left(\frac{\sin 2\theta}{\sin \theta}\right)^4 \,d\theta\)
   ​\(= \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 2\theta}{8 \sin^4 \theta} \,d\theta = \int_{0}^{\frac{\pi}{2}} \frac{2^5 \sin^5 \theta \cos^5 \theta}{8 \sin^4 \theta} \,d\theta = 4 \int_{0}^{\frac{\pi}{2}} \sin \theta \cos^5 \theta \,d\theta\)
   Let \(t = \cos \theta\). Then \(dt = -\sin \theta \,d\theta\).
   ​\(= 4 \int_{1}^{0} t^5 (-dt) = 4 \int_{0}^{1} t^5 \,dt = 4 \left( \frac{1}{6} t^6 \Big|_{0}^{1} \right) = 4 \times \frac{1}{6} = \frac{2}{3}\)

2. (20 points)
   Let \(S\) be the triangle with vertices \((0,0), (1,0)\) and \((0,1)\). Let \(f(x,y) = e^{\frac{x-y}{x+y}}\). Evaluate the integral \(\int_S f dA\) by making the change of variables \(u = x-y, v = x+y\).

   ​

   We know \(x = \frac{v+u}{2}\), \(y = \frac{v-u}{2}\), then \(g(u,v) = (\frac{v+u}{2}, \frac{v-u}{2})\).
   When \(x=0\), \(v=-u\). When \(y=0\), \(v=u\), when \(y=-x+1\), \(v=1\).
   Then \(\int_S f dA = \int_0^1 \int_{-v}^v e^{\frac{u}{v}} \cdot |J_g(u,v)| \,du\,dv\)

   \(|J_g(u,v)| = det\begin{pmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{pmatrix} = det\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)

   Then \(\int_S f dA = \int_0^1 \int_{-v}^v e^{\frac{u}{v}} \cdot \frac{1}{2} \,du\,dv = \frac{1}{2} \int_0^1 (e^{\frac{u}{v}} \cdot v \Big|_{-v}^v) \,dv\)
   ​\(= \frac{1}{2} \int_0^1 (e \cdot v - e^{-1} \cdot v) \,dv = \frac{1}{2} (\frac{e}{2} \cdot v^2 - \frac{e^{-1}}{2} \cdot v^2 \Big|_0^1) = \frac{1}{2} (\frac{e}{2} - \frac{e^{-1}}{2}) = \frac{1}{4}(e-e^{-1})\)

3. (30 points)
   Compute the following integrals with the suggested change of variables.
   (a) \(\int_0^2 \int_{y/2}^{(y+4)/2} y^3(2x-y)e^{(2x-y)^2} dxdy\), using the transformation \(x = u + \frac{1}{2}v, y = v\).

   When \(x = \frac{y+4}{2}\), \(u=2\). When \(x = \frac{y}{2}\), \(u=0\). When \(y=2\), \(v=2\). When \(y=0\), \(v=0\).
   Then \(\int_0^2 \int_{y/2}^{(y+4)/2} y^3(2x-y)e^{(2x-y)^2} dxdy = \int_0^2 \int_0^2 v^3 \cdot 2u \cdot e^{4u^2} |J_g| dvdu\).
   ​\(|J_g| = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det \begin{pmatrix} 1 & \frac{1}{2} \\ 0 & 1 \end{pmatrix} = 1\).
   Then \(= \int_0^2 \int_0^2 v^3 \cdot 2u \cdot e^{4u^2} dvdu = \int_0^2 \frac{1}{4} \cdot 2u \cdot e^{4u^2} v^4 \Big|_0^2 du = \int_0^2 \frac{u}{2} e^{4u^2} \cdot 16 du\).
   ​\(= \int_{0}^{2} 8u \cdot e^{4u^2}du = \int_{0}^{4} e^{4t}dt = 4 \int_{0}^{4} e^{4t} dt = e^{4t}\Big|_{0}^{4} = e^{16}-1\).

   (b) \(\iint_S (x^2-y^2) dA\), where \(S\) is the region bounded by the curves \(y-x=1, y-x=0, xy=1\) and \(xy=2\), with the change of coordinates \(u=y-x, v=xy\).

   When \(xy=1, v=1\), When \(xy=2, v=2\). When \(y-x=0, u=0\), When \(y-x=1, u=1\).

   Then \(u=y-x\), \(v=xy\), \(x=\frac{\sqrt{u^{2}+4v}-u}{2},y=\frac{\sqrt{u^{2}+4v}+u}{2}\)

   Then \(|J_{g}|=det \begin{pmatrix} -\frac{1}{2}{+}\frac{u}{2\sqrt{u^{2}+4v}} & \frac{2}{\sqrt{u^{2}+4v}} \\ \frac{1}{2}{+}\frac{u}{2\sqrt{u^{2}+4v}} & \frac{2}{\sqrt{u^{2}+4v}} \end{pmatrix}=\left|-\frac{1}{\sqrt{u^{2}+4v}}\right|=\frac{1}{\sqrt{u^{2}+4v}}\)

   Then \(\int_{S}fdA=\int_{0}^{1}\int_{1}^{2}-u\sqrt{u^{2}+4v}\times\frac{1}{\sqrt{u^{2}+4v}} dvdu=\int_{0}^{1}\int_{1}^{2}-udvdu\)
   ​\(= \int_0^1 -u du = -\frac{1}{2}u^2 \Big|_0^1 = -\frac{1}{2}\)