MHB Evaluating a double integral in polar coordinates

[skate_nerd]

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]

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 [math] \displaystyle 2\pi [/math], which means your [math] \displaystyle \theta [/math] bounds are actually 0 to [math]\displaystyle 2\pi [/math].

 

[chisigma]

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

$\chi$ $\sigma$

 

[skate_nerd]

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!