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

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- Jan 1st 2009, 06:05 PMmitch_nufcdouble integration with polar coordinates
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 - Jan 1st 2009, 06:45 PMMush
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 $\displaystyle y = \sqrt{a^2-x^2} $

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

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

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

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

$\displaystyle 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 $

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

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