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 September 11th, 2018, 03:58 AM #1 Newbie   Joined: Sep 2018 From: Macedonia [FYROM] Posts: 1 Thanks: 0 How to solve this double integral I've got a final exam tomorrow and i am stuck on this exercise. Could you help me? The Exercise: S is an area enclosed with a circle with radius 'a', the circle touches both coordinate axes and is located in the 2nd quadrant. Solve  September 11th, 2018, 12:05 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 translate it to polar coordinates to obtain $\displaystyle \int_{\frac \pi 2}^\pi \int_0^a~\dfrac{r dr d\theta}{\sqrt{2a+r\cos(\theta)}}$ Mathematica returns $\dfrac{2}{3} a^{3/2} \left(-2 \sqrt{3} K\left(\frac{2}{3}\right)+\sqrt{2}+2 \sqrt{3} F\left(\frac{\pi }{4},\frac{2}{3}\right)+4 \sqrt{3} E\left(\frac{2}{3}\right)-4 \sqrt{3} E\left(\frac{\pi }{4},\frac{2}{3}\right)\right)$ where $K,~F,~E$ are all elliptic functions The integral approximately equals $0.632837 a^{3/2}$ September 11th, 2018, 01:19 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 How does that region "touch both axes"? September 11th, 2018, 01:44 PM   #4
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 Originally Posted by skipjack How does that region "touch both axes"?
pretty sure that's the 1/4 disk of radius a that lies in the second quadrant September 11th, 2018, 01:58 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 The entire disk should be in the 2nd quadrant. September 11th, 2018, 02:18 PM   #6
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 Originally Posted by skipjack The entire disk should be in the 2nd quadrant.
you're right September 11th, 2018, 02:33 PM #7 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 ok.. first we have to translate the coordinate system so that $\left(-\dfrac a 2, \dfrac a 2\right) \to (0,0)$ $x \to x+\dfrac a 2,~y \to y-\dfrac a 2$ this gets us the integral $\displaystyle \int \int_S \dfrac{1}{\sqrt{2a+\left(x+\frac a 2\right)}}$ Now $S$ is the disk of radius $\dfrac a 2$ centered at the origin. Proceeding as before but with new limits of integration we get $\displaystyle \int_0^{2\pi} \int_0^{\frac a 2}~\dfrac{r dr d\theta}{\sqrt{2a+r\cos(\theta)+\frac a 2}} =\int_0^{2\pi} \int_0^{\frac a 2}~\dfrac{r dr d\theta}{\sqrt{\frac 5 2 a+r\cos(\theta)}}$ Mathematica returns $\displaystyle \int_0^{2\pi} \int_0^{\frac a 2}~\dfrac{r dr d\theta}{\sqrt{\frac 5 2 a+r\cos(\theta)}} = \dfrac{4 \left(5 E\left(\dfrac{1}{3}\right)-4 K\left(\dfrac{1}{3}\right)\right)}{\sqrt{3}}$ I'm not crazy about this answer because of it's lack of dependence on $a$ What do you think SkipJack? September 11th, 2018, 10:59 PM #8 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 The translation would be $\displaystyle x \to x - a,~y \to y + a$, so that $\displaystyle (-a, a) \to (0,0)$. That would lead to $\displaystyle \int_0^{2\pi} \!\int_0^a \frac{r dr d\theta}{\sqrt{a + r\cos(\theta)}}$. September 13th, 2018, 07:25 AM #9 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Take new coordinate system at center of circle. x'=x+a y'=y-a Drop primes and problem in new coordinate system becomes: $\displaystyle I=\int \int \frac{1}{\sqrt{a+x}}dxdy$ over disc $\displaystyle x^{2}+y^{2}=a^{2}$ $\displaystyle I=\int \frac{1}{\sqrt{a+x}}\int dydx$ $\displaystyle \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}dy=2\sqrt{a^{2}-x^{2}}$ $\displaystyle I=2\int_{-a}^{a}\frac{\sqrt{a^{2}-x^{2}}}{\sqrt{a+x}}dx=2\int_{-a}^{a}\sqrt{a-x}dx= \frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=-\frac{8}{3}a$ EDIT: Acknowledgement. Equations typed with: https://www.codecogs.com/latex/eqneditor.php Last edited by zylo; September 13th, 2018 at 07:36 AM. September 13th, 2018, 12:02 PM   #10
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 Originally Posted by zylo $\displaystyle 2\int_{-a}^{a}\sqrt{a-x}dx= \frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=-\frac{8}{3}a$ Tags double, integral, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DakshD Calculus 1 November 8th, 2016 12:38 PM Jhenrique Calculus 5 June 30th, 2015 04:45 PM mathbalarka Calculus 9 May 28th, 2012 06:32 AM maximus101 Calculus 0 March 4th, 2011 02:31 AM maximus101 Algebra 0 December 31st, 1969 04:00 PM

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