Transform to polar and calculate

[bamptom]

Hi helpers,

I have a homework question which asking me to Transform to polar and calculate //(3y-x)dydx for 0<x<3, -sqrt(9-x^2) < y < sqrt(9-x^2)

I think i have defined the region by 3pi/2<theta<pi/2 and 0<r<3

I am unsure how to substitute 3y-x into the //f(rcostheta,rsintheta)rdrdtheta equation

If anybody could point me in the right direction it would be very much appreciated

 

[galactus]

\(\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{3}r(3rsin{\theta}-rcos{\theta})drd{\theta}\)

 

[bamptom]

Thankyou very very much that confirms it for me