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September 11th, 2018, 02:58 AM   #1
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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
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September 11th, 2018, 11:05 AM   #2
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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 $0.632837 a^{3/2}$
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September 11th, 2018, 12:19 PM   #3
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How does that region "touch both axes"?
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September 11th, 2018, 12:44 PM   #4
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Quote:
Originally Posted by skipjack View Post
How does that region "touch both axes"?
pretty sure that's the 1/4 disk of radius a that lies in the second quadrant
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September 11th, 2018, 12:58 PM   #5
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The entire disk should be in the 2nd quadrant.
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September 11th, 2018, 01:18 PM   #6
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Quote:
Originally Posted by skipjack View Post
The entire disk should be in the 2nd quadrant.
you're right
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September 11th, 2018, 01:33 PM   #7
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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?
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September 11th, 2018, 09:59 PM   #8
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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 $\displaystyle \int_0^{2\pi} \!\int_0^a \frac{r dr d\theta}{\sqrt{a + r\cos(\theta)}}$.
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September 13th, 2018, 06:25 AM   #9
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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 06:36 AM.
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September 13th, 2018, 11:02 AM   #10
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Quote:
Originally Posted by zylo View Post
$\displaystyle 2\int_{-a}^{a}\sqrt{a-x}dx=
\frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=-\frac{8}{3}a $
Correction: Missing minus sign in second term which makes answer positive as it should be.
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