00:01
In this question, we want to evaluate the integral from 0 to 2 pi of the integral from 0 to 3 of the integral from 0 to z over 3 of r cubed dr dz d theta.
00:11
So we just work from the inside out.
00:12
We start by finding the antiderivative of r cubed with respect to r.
00:18
What is that? so we have integral 0 to 2 pi of the integral 0 to 3 of r to the fourth over 4.
00:28
And then this is being evaluated from 0 to z over 3 dz d theta.
00:37
So what do we have? integral 0 to 2 pi integral 0 to 3.
00:44
When i plug in z over 3, i'm getting z to the fourth over 3 to the fourth.
00:51
3 to the fourth is 81.
00:54
This is over 4.
00:55
All of this dz d theta.
00:59
I have a 1 over 324 that i can pull through so that i have 1 over 324 times the integral from 0 to 2 pi of the integral from 0 to 3 of z to the fourth dz d theta.
01:20
Now, my antiderivative of z to the fourth is z to the fifth over 5.
01:25
So 1 over 324 integral 0 to 2 pi of z to the fifth over 5.
01:34
Evaluate this from 0 to 3 d theta.
01:38
This is giving us 1 over 324 times the integral from 0 to 2 pi...