double integration with polar coordinates

mitch_nufc · #1 $Old$ January 1st, 2009, 08:05 PM

By changing to polar coordinates, evaluate the integral (where a>0):

int[a,0]int[sqrt(a^2 - x^2), 0](x^2 + y^2)dydx

Mush · #2 $Old$ January 1st, 2009, 08:45 PM

May aswell post here!

Draw a picture of the region of the limits.

The lower limits are x = 0 and y = 0, which tells you that both x > 0 and y > 0, which means you're dealing with the 1st quadrant of the cartesian axis system.

Sketching the upper y limit $y = \sqrt{a^2-x^2}$

gives: $y^2 + x^2 = a^2$

Circle with centre (0,0), radius a.

So from that sketch you should see that, in polar, the $\theta$ limits range from $0 \to \frac{\pi}{2}$ .

And the radius ranges from $0 \to a$ . Which converts to:

$I = \int_0^{\frac{\pi}{2}} \int_0^{a} ((rcos(\theta))^2+(rsin(\theta))^2).|\frac{d(x,y)}{d(r,\theta)}| dr d\theta$

$|\frac{d(x,y)}{d(r,\theta)}| = r$

$I = \int_0^{\frac{\pi}{2}} \int_0^{a} r^3 dr d\theta$