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Evaluating a double integral in polar coordinates

skatenerd

Active member
Oct 3, 2012
114
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to \(d\theta\) and I wasn't entirely sure how to integrate that so I thought it might just be \(\pi\).
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius \(a\) and the answer has nothing to do with \(a\). If someone could let me know where I went wrong that would be great.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Re: evaluating a double integral in polar coordinates

Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually \(\displaystyle \displaystyle 2\pi \), which means your \(\displaystyle \displaystyle \theta \) bounds are actually 0 to \(\displaystyle \displaystyle 2\pi \).
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: evaluating a double integral in polar coordinates

I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to \(d\theta\) and I wasn't entirely sure how to integrate that so I thought it might just be \(\pi\).
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius \(a\) and the answer has nothing to do with \(a\). If someone could let me know where I went wrong that would be great.
The bounds of the inner integral are 0 and a, not -a and a so that is...

$\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1)

Kind regards

$\chi$ $\sigma$
 

skatenerd

Active member
Oct 3, 2012
114
Re: evaluating a double integral in polar coordinates

Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it!