00:01
In this question we are asked to evaluate this integration over the region r by changing to polar coordinates.
00:07
So let's start to solve this problem.
00:09
To change this integral into the polar coordinates, we substitute x equals to r, cos theta, y equals to r, sine theta, d x into d, y is equal to r, dr d theta.
00:26
Over the region r, small r is varying from, 0 to radius of this circle is equal to 9 and theta is varying from in the first quadrant theta is varying from 0 to pi over 2 d x into dy x into d y is equal to d a now after putting all these values here this integration becomes y is equal to r sine theta t to the power r cost theta r d r d theta r is varying from 0 to 9 theta is varying from 0 to pi over 2 r into r is equal to r square sine theta a to the power r cost theta d r d theta 0 to pi over 2 r r square, sine theta, e to the power r, cos theta, d theta.
01:53
Now here we change the order of this integration.
01:57
Here both limits are constant.
01:59
That's why we can write this integration is equal to double integration.
02:04
Theta is varying from 0 to pi over to 2, r is varying from 0 to 9.
02:11
R squared, sine theta e to the power r, cos theta, d theta into dr.
02:24
Integration 0 to 9, r squared, integration 0 to pi over 2, sine theta, e to the power r, cos theta, d theta, dr...