# Thread: Convert double integral to polar

1. ## Convert double integral to polar

Hi Guys / girls

I am going through my home work for uni and i have to calculate the integral of the following by transforming to polar coordinates. This fine except i just cant get my head around how to do this.

The problem is -//(3y-x)dydx 0<x<3, -sqrt(9-x^2) < y < sqrt(9-x^2)

2. Originally Posted by bamptom
Hi Guys / girls

I am going through my home work for uni and i have to calculate the integral of the following by transforming to polar coordinates. This fine except i just cant get my head around how to do this.

The problem is -//(3y-x)dydx 0<x<3, -sqrt(9-x^2) < y < sqrt(9-x^2)

Have you drawn the region of integration?

Note: $-\sqrt{9-x^2} < y < \sqrt{9-x^2}$ is equivalent to the interior of the circle $x^2 + y^2 = 9$.
$0 < x < 3$ means you want the right half of this circle ......

Can you define this region using polar coordinates? That will give you the integral terminals.

And you know $dx \, dy \rightarrow r \, dr\, d\theta$, right?

3. Are they 0<Theta<pi and 0<r<3 ??

And does (3y-x)dydx >>>>> // (rcostheta + 3rsintheta)rdrdtheta

I really do appreciate your help!

4. ## ??

Can anyone tell me if im on the right track here?

5. Originally Posted by bamptom
Are they 0<Theta<pi and 0<r<3 ??

[snip]
No. Have you drawn the region like I suggested? What's the polar coordinates of (0, -3) and (0, 3) .....? Will $0 < \theta < \pi$ give you those points .....?

Originally Posted by bamptom
[snip]
And does (3y-x)dydx >>>>> // (rcostheta + 3rsintheta)rdrdtheta

I really do appreciate your help!
No. How can you substitute $x = r \cos \theta$ and $y = r \sin \theta$ into $3y - x$ and get $r \cos \theta + 3r \sin \theta$ ....?

By the way, don't bump.