Thread: Integrate using polar coordinates

Evaluate the iterated integral by converting to polar coordinates.

integral(0,a)integral(-sqrt(a^2-y^2), 0) x^2y dxdy

I transfered the x^2y into polar coordinates:
x= rcos(theta) , y= rsin(theta)
(rcos(theta)^2(rsin(theta))r drd(theta)

but, i'm not sure how to transform the integral boundaries into polar coordiantes. Usually the first one would be like 0-pi but i'm not sure with the a being there. The a is throwing me off.

Originally Posted by cul8er

Evaluate the iterated integral by converting to polar coordinates.

integral(0,a)integral(-sqrt(a^2-y^2), 0) x^2y dxdy

I transfered the x^2y into polar coordinates:
x= rcos(theta) , y= rsin(theta)
(rcos(theta)^2(rsin(theta))r drd(theta)

but, i'm not sure how to transform the integral boundaries into polar coordiantes. Usually the first one would be like 0-pi but i'm not sure with the a being there. The a is throwing me off.

First, let's get an idea of what the domain being integrated over is for the original function:

D = {(x,y) | 0 <= y <= a, -sqrt(a^2 - y^2) <= x <= 0}

x = -sqrt(a^2 - y^2) --> this is the left half of a circle with radius a
y = 0 -> a --> on the left half of the circle, this is the upper portion

From these, I can conclude that we are dealing with 1/4 of a circle. To be precise, we are dealing with the 1/4 of the circle that is in the 2nd quadrant. From this, I can conclude that the polar domain becomes:

D = {(r,theta) | 0 <= r <= a, pi/2 <= theta <= pi}

Can you do the problem from here?

Ok, so here is what i got:

int(0,a)int(pi/2, pi) (r^2cos(theta)^2)(rsin(theta))
int(0,a)r^4 dr = r^5/5 [0,a] = a^5/5
int(pi/2, pi) (cos(theta)^2)(sin(theta)) = -1/3cos(theta)^3 [pi/2, pi] = 1/3
1/3 * a^5/5
= a^5/15

Is that correct?
I wasn't sure if r^4 is correct. I was separating the r's from the cos and sin, and combined with the other r, i got r^4.
And i also wasn't sure if the integration of (cos(theta)^2)(sin(theta)) was correct.