Double Integral Using Polar Coordinates PDF Math 3 2020

# Double Integral Using Polar Coordinates ## Double Integral Using Polar Coordinates * Double Integral * Polar Coordinates **Math 3** **2020** **Double Integral Using Polar Coordinates** $\iint_R f(x,y)dA \longrightarrow \iint_R f(r, \theta) rdrd\theta$ $x = rcos\theta$ $y = rsin\theta$ $x^2...

# Double Integral Using Polar Coordinates ## Double Integral Using Polar Coordinates * Double Integral * Polar Coordinates **Math 3** **2020** **Double Integral Using Polar Coordinates** $\iint_R f(x,y)dA \longrightarrow \iint_R f(r, \theta) rdrd\theta$ $x = rcos\theta$ $y = rsin\theta$ $x^2 + y^2 = r^2$ $\theta = tan^{-1}\frac{y}{x}$ $dA = dxdy = rdrd\theta$ **Important Notes** 1. **For** $x^2 + y^2 = a^2$ * **Equation of circle with origin as the center** * **Radius is (a)** * **r = a** > **Complete Circle** > $0 \le \theta \le 2\pi$ **Upper Half** > $0 \le \theta \le \pi$ > $0 \le r \le a$ **Lower Half** > $\pi \le \theta \le 2\pi$ or $\pi \le \theta \le 0$ > $0 \le r \le a$ **First Quadrant** > $0 \le \theta \le \frac{\pi}{2}$ > $0 \le r \le a$ **Second Quadrant** > $\frac{\pi}{2} \le \theta \le \pi$ > $0 \le r \le b$ > $a \le r \le b$ 2. * **Equation of line** $y = x$ * **Area** under the line * **Limits of Integration** * $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ 3. * **Equation of line** $y = x$ * **Area** under the line * **Limits of Integration** * $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$ 4. * **Equation of line** $y = x$ * **Equation of line** $y = -x $ * **Area** under the line * **Limits of Integration** * $\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}$ **Example 1** **Evaluate the double integral:** $\int \int_R x^2y^2 dA $ **where R is the region in the first quadrant bounded by the circles:** $x^2+y^2 = 1$ and $y^2 + x^2 = 4$ $\int \int_R x^2y^2 dA = \int \int_R (rcos\theta)^2 (rsin\theta)^2 rdrd\theta = \int \int_R r^5 cos^2 \theta sin^2 \theta drd\theta$ $\int_0^{\pi/2} \int_1^2 r^5 cos^2 \theta sin^2 \theta drd\theta = \int_0^{\pi/2} [ \frac{r^6}{6} cos^2 \theta sin^2 \theta ]_1^2 d\theta$ $\int_0^{\pi/2} \[ \frac{2^6}{6} cos^2 \theta sin^2 \theta - \frac{1^6}{6} cos^2 \theta sin^2 \theta \] d\theta$ $\int_0^{\pi/2} \frac{63}{6} cos^2 \theta sin^2 \theta d\theta = \frac{63}{6} \int_0^{\pi/2} cos^2 \theta sin^2 \theta d\theta$ $\frac{63}{6} \int_0^{\pi/2} \frac{1}{4}sin^2 (2\theta) d\theta = \frac{63}{24} \int_0^{\pi/2} \frac{1}{2} (1 - cos(4\theta)) d\theta$ $\frac{63}{48} [ \theta - \frac{sin(4\theta)}{2}\ ]_0^{\pi/2} = \frac{63}{48} [ (\frac{\pi}{2} - \frac{sin(2\pi)}{4}) - (0 - \frac{sin(0)}{4})]$ $\frac{63}{48} ( \frac{\pi}{2} - 0) = \boxed{\frac{63\pi}{96}}$ **Example 2** **Find:** $\int_0^1 \int_0^{\sqrt{1 - y^2}} e^{x^2+y^2} dxdy$ $\int_0^1 \int_0^{\sqrt{1 - y^2}} e^{x^2+y^2} dxdy = \int_0^{\pi/2} \int_0^1 e^{r^2} rdrd\theta$ $\frac{1}{2} \int_0^{\pi/2} \int_0^1 e^{u} du d\theta = \frac{1}{2} \int_0^{\pi/2} [e^{u} ]_0^1 d\theta$ $\frac{1}{2} \int_0^{\pi/2} [ e^1 - e^0 ] d\theta = \frac{1}{2} \int_0^{\pi/2} (e - 1) d\theta$ $\frac{1}{2} [(e - 1)\theta ]_0^{\pi/2} = \frac{1}{2}[(e - 1)\frac{\pi}{2} - (e-1)(0)] = \boxed{\frac{\pi}{4} (e - 1)}$ **Example 3** **Evaluate:** $\int_0^2 \int_0^{\sqrt{8-x^2}} \frac{1}{5+x^2+y^2} dydx$ $\int_0^2 \int_0^{\sqrt{8-x^2}} \frac{1}{5+x^2+y^2} dydx = \int_{\pi/4}^{\pi/2} \int_2^{\sqrt{8}} \frac{1}{5+r^2} rdrd\theta$ $\int_{\pi/4}^{\pi/2} \int_2^{\sqrt{8}} \frac{r}{5+r^2} drd\theta = \frac{1}{2} \int_{\pi/4}^{\pi/2} \[ ln(5+r^2) \]_2^{\sqrt{8}} d\theta$ $\frac{1}{2} \int_{\pi/4}^{\pi/2} \[ln(5 + 8) - ln(5 + 4) \] d\theta = \frac{1}{2} \int_{\pi/4}^{\pi/2} \[ln(13) - ln(9) \] d\theta$ $\frac{1}{2} \int_{\pi/4}^{\pi/2} ln(\frac{13}{9}) d\theta = \frac{1}{2} ln(\frac{13}{9}) \[ \theta \]_{\pi/4}^{\pi/2} = \frac{1}{2} ln(\frac{13}{9}) (\frac{\pi}{2} - \frac{\pi}{4})$ $\boxed{\frac{\pi}{8} ln(\frac{13}{9})}$ **Example 4** **Evaluate:** $\int_0^{2\sqrt{2}} \int_0^{\sqrt{16-x^2}} \frac{1}{\sqrt{9+x^2+y^2}} dydx$ $\int_0^{2\sqrt{2}} \int_0^{\sqrt{16-x^2}} \frac{1}{\sqrt{9+x^2+y^2}} dydx = \int_{\pi/4}^{\pi/2} \int_0^4 \frac{r}{\sqrt{9+r^2}} drd\theta$ $\int_{\pi/4}^{\pi/2} \int_0^4 (9+r^2)^{-1/2} rdrd\theta = \int_{\pi/4}^{\pi/2} \[ \sqrt{9+r^2} \]_0^4 d\theta$ $\int_{\pi/4}^{\pi/2} \[\sqrt{25} - \sqrt{9} \] d\theta = \int_{\pi/4}^{\pi/2} (5 - 3) d\theta = \int_{\pi/4}^{\pi/2} 2 d\theta$ $\boxed{\frac{\pi}{2}}$ **Example 5** **Evaluate** $\int_0^a \int_0^{\sqrt{a^2-x^2}} \frac{1}{\sqrt{x^2 + y^2}} dydx$ $\int_0^a \int_0^{\sqrt{a^2-x^2}} \frac{1}{\sqrt{x^2 + y^2}} dydx = \int_0^{\pi/2} \int_0^a \frac{r}{r} drd\theta$ $\int_0^{\pi/2} \int_0^a drd\theta = \int_0^{\pi/2} \[r \]_0^a d\theta$ $\int_0^{\pi/2} (a - 0) d\theta = \int_0^{\pi/2} a d\theta = a \[\theta\]_0^{\pi/2} = a(\frac{\pi}{2} - 0) = \boxed{\frac{a\pi}{2}}$ **Example 6** **Evaluate:** $\int_{-\pi}^{\pi} \int_0^{\sqrt{\pi - x^2}} sin(x^2 + y^2) dydx$ $\int_{-\pi}^{\pi} \int_0^{\sqrt{\pi - x^2}} sin(x^2 + y^2) dydx = \int_0^{\pi} \int_0^{\sqrt{\pi}} sin(r^2) rdrd\theta$ $\frac{1}{2} \int_0^{\pi} \int_0^{\pi} sin(u) du d\theta = \frac{1}{2} \int_0^{\pi} \[ -cos(u) \]_0^{\pi} d\theta$ $\frac{1}{2} \int_0^{\pi} [-cos(\pi) - (-cos(0))]d\theta = -\frac{1}{2} \int_0^{\pi} (-1 - (-1)) d\theta$ $-\frac{1}{2} \int_0^{\pi} 0 d\theta = \boxed{0 }$ **Example 7** **Evaluate:** $\int_0^1 \int_0^{\sqrt{2-y^2}} (x+y) dxdy$ $\int_0^1 \int_0^{\sqrt{2-y^2}} (x+y) dxdy = \int_0^{\pi/4} \int_0^{\sqrt{2}} (rcos\theta + rsin \theta) rdrd\theta$ $\int_0^{\pi/4} \int_0^{\sqrt{2}} (r^2 cos\theta + r^2 sin\theta) drd\theta = \int_0^{\pi/4} [ \frac{r^3}{3} cos\theta + \frac{r^3}{3} sin\theta ]_0^{\sqrt{2}} d\theta$ $\int_0^{\pi/4} [(\frac{(\sqrt{2})^3}{3} cos\theta + \frac{(\sqrt{2})^3}{3} sin\theta) - 0 + 0]d\theta$ $\int_0^{\pi/4} (\frac{2\sqrt{2}}{3} cos\theta + \frac{2\sqrt{2}}{3} sin\theta) d\theta = [\frac{2\sqrt{2}}{3} sin \theta - \frac{2\sqrt{2}}{3} cos \theta ]_0^{\pi/4}$ $\frac{2\sqrt{2}}{3} ( sin(\frac{\pi}{4} - cos (\frac{\pi}{4})) - \frac{2\sqrt{2}}{3} (sin(0) - cos(0))]$ $\frac{2\sqrt{2}}{3} ( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} ) - \frac{2\sqrt{2}}{3} (0 - 1) = \boxed{\frac{2\sqrt{2}}{3}}$ **Example 8** **Evaluate** $\iint_R \sqrt{x^2 + y^2 - 1} dA$, where R is the region bounded by $x^2 + y^2 = 1$, $x^2 + y^2 = 4$, $y = x$, $y = -x$ in the upper half plane. $\iint \sqrt{x^2 + y^2 - 1} dA = \int_{\pi/4}^{3\pi/4} \int_1^2 \sqrt{r^2 - 1} rdrd\theta$ $\int_{\pi/4}^{3\pi/4} \int_1^2 (r^2 - 1)^{1/2} rdrd\theta = \int_{\pi/4}^{3\pi/4} \int_1^2 (r^2 - 1)^{1/2} rdrd\theta$ $\int_{\pi/4}^{3\pi/4} \int_1^2 u^{1/2} \frac{1}{2} du d\theta = \frac{1}{2} \int_{\pi/4}^{3\pi/4} [\frac{2}{3} u^{\frac{3}{2}}]_1^2 d\theta$ $\frac{1}{3} \int_{\pi/4}^{3\pi/4} [(2)^{\frac{3}{2}} - 1] d\theta = \int_{\pi/4}^{3\pi/4} [(2)^{\frac{3}{2}} - 1] d\theta$ $\frac{1}{3} [(2)^{\frac{3}{2}} - 1] [\theta ]_{\pi/4}^{3\pi/4} = \frac{1}{3} [(2)^{\frac{3}{2}} - 1] (\frac{3 \pi}{4} - \frac{\pi}{4})$ $\boxed{\frac{1}{6} \sqrt{27} (\frac{3 \pi}{4} - \frac{\pi}{4})}$ **Example 9** **Evaluate:** $\int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \frac{2}{(x^2 + y^2 + 1)^3} dydx$ $\int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \frac{2}{(x^2 + y^2 + 1)^3} dydx = \int_0^{2\pi} \int_0^1 \frac{2}{(r^2 + 1)^3} rdrd\theta$ $\int_0^{2\pi} \int_0^1 2(r^2 + 1)^{-3} rdrd\theta = 2\int_0^{2\pi} \int_0^1 (r^2 + 1)^{-3} rdrd\theta$ $2 \int_0^{2\pi} \int_0^1 u^{-3} \frac{1}{2}du d\theta = \int_0^{2\pi} \int_0^1 u^{-3} du d\theta$ $\int_0^{2\pi} [-\frac{1}{2} u^{-2}]_0^1 d\theta = \int_0^{2\pi} [-\frac{1}{2} (1)^{-2} - (-\frac{1}{2} (0)^{-2}] d\theta$ $\int_0^{2\pi} [ -\frac{1}{2} ] d\theta = -\frac{1}{2} [\theta ]_0^{2\pi} = -\frac{1}{2} (2\pi - 0)$ $\boxed{-\pi}$ **Example 10** **Evaluate:** $\int_0^\infty \int_0^\infty \frac{-y}{e^x} dxdy$ $\int_0^\infty \int_0^\infty \frac{-y}{e^x} dxdy = \int_0^\infty \int_0^\infty -ye^{-x} dxdy$ $\int_0^\infty -y \[-e^{-x} \]_0^\infty dy = \int_0^\infty -y [0 - (-1)]dy$ $\boxed{1}$