Question

Evaluate by using polar coordinates. ∫-2 to 2 ∫ 0 to sqrt(9-y^2) f(x,y)=sqrt(x^2+y^2) dxdy. I'm having trouble defining the bounds for r when turned into polar coordinates.

Answer 1

\[\int_{-2}^2\int_0^\sqrt{9-y^2}\sqrt{x^2+y^2}dxdy\]

Answer 2

\[x =r cos \theta\rightarrow x^2 = r^2cos^2\theta\\y=rsin\theta\rightarrow y^2=r^2sin^2\theta\\--------------\\x^2+y^2=r^2\\\sqrt{x^2+y^2}=r\]
so the integrand turns to \(r^2drd\theta\) I think you know the reason why it is, right? (Jacobian number)
Now, the bounds: let see the bound from dx , it says x goes from 0 \(\rightarrow \sqrt{9-y^2}\). That means \(x^2 = 9 -y^2\rightarrow x^2+y^2=3^2\), so r goes from 0 \(\rightarrow 3\)

Answer 3

then , look at the limits of y, it goes from -2 to 2 . It means \(\theta\) goes from \(-\pi/2 \rightarrow \pi/2\)