# Evaluate the integral by converting to polar coordinates

## Homework Statement

Evaluate the integral
integral from 0 to 1 of (integral from y to sqrt(2-y^2) of (3(x-y))dxdy)
by converting to polar coordinates.

x = rcos(theta)
y = rsin(theta)

## The Attempt at a Solution

By drawing a picture of the bounds, I concluded that r goes from 0 to sqrt(2) and theta goes from pi/4 to pi/2.
I tried to integrate like this:
integral from 0 to sqrt(2) of 3(rcos(theta) - rsin(theta) r dr dtheta
= integral from 0 to sqrt(2) of 3r^2 (cos(theta) - sin(theta)) dr dtheta
= r^3(cos(theta) - sin(theta)) where r is from 0 to sqrt(2)
=2sqrt(2) * integral from pi/4 to pi/2 of (cos(theta)-sin(theta)) dtheta
=2sqrt(2) * (sin(theta)+cos(theta)) with theta from pi/4 to pi/2
=2sqrt(2) *((1+0)-((sqrt(2)/2) + (sqrt(2)/2))
=2sqrt(2) *(1-sqrt(2))
=2sqrt(2) - 4

The actual answer is 4 - 2sqrt(2) so I missed a sign somewhere, but I cannot find it. Sorry if it's difficult to read. Any help would be greatly appreciated.

Mark44
Mentor

## Homework Statement

Evaluate the integral
integral from 0 to 1 of (integral from y to sqrt(2-y^2) of (3(x-y))dxdy)
by converting to polar coordinates.

x = rcos(theta)
y = rsin(theta)

## The Attempt at a Solution

By drawing a picture of the bounds, I concluded that r goes from 0 to sqrt(2) and theta goes from pi/4 to pi/2.
Here's your problem. The region of integration is the sector of a circle that extends from theta = 0 to theta = pi/4. The radius is as you have it.
I tried to integrate like this:
integral from 0 to sqrt(2) of 3(rcos(theta) - rsin(theta) r dr dtheta
= integral from 0 to sqrt(2) of 3r^2 (cos(theta) - sin(theta)) dr dtheta
= r^3(cos(theta) - sin(theta)) where r is from 0 to sqrt(2)
=2sqrt(2) * integral from pi/4 to pi/2 of (cos(theta)-sin(theta)) dtheta
=2sqrt(2) * (sin(theta)+cos(theta)) with theta from pi/4 to pi/2
=2sqrt(2) *((1+0)-((sqrt(2)/2) + (sqrt(2)/2))
=2sqrt(2) *(1-sqrt(2))
=2sqrt(2) - 4

The actual answer is 4 - 2sqrt(2) so I missed a sign somewhere, but I cannot find it. Sorry if it's difficult to read. Any help would be greatly appreciated.

That solves the problem! Thank you very much!

Last edited:
Mark44
Mentor
Sure, you're welcome. As you found out, it's very important to have an accurate understanding of the region of integration. In many problems of this type, the hardest part can often be representing the region of integration in a different coordinate system.