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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: 2,311 Thanks: 1223 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: 20,285 Thanks: 1967 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: 20,285 Thanks: 1967 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: 2,311 Thanks: 1223 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: 20,285 Thanks: 1967 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: 124 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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Quote:
 Originally Posted by zylo $\displaystyle 2\int_{-a}^{a}\sqrt{a-x}dx= \frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=-\frac{8}{3}a$

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