Thread: 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)

Thanks in advance

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)

Thanks in advance

Have you drawn the region of integration?

Note: is equivalent to the interior of the circle .
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 , right?

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

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

I really do appreciate your help!

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

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 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 and into and get ....?

By the way, don't bump.