To evaluate double integral using Polar coordinates

Question

$$ \iint 4y - (x^2 + y^2) dxdy$$ over region $x^2 + (y-2)^2 = 2^2$. Basically I tried to solve this using twoo methods but I am getting different results.I want to know whether my method are correct (not computation part but otherwise setting up of integral)

I put $x = r \cos\theta$ and $y = r \sin\theta$ and my $r$ goes from $0$ to $4\sin\theta$ and $\theta $ goes from $0$ to $\pi$
I use change of variables by putting $x = r \cos\theta$ and $y = 2 + 2r \sin\theta$. and i get $J= 4r\,drd\theta$. o integral limits are $r$ goes from $0$ to $1$ and theta goes from $0$ to $2\pi$

My question is if my limits and integrand are correct in both methods or not

I think they are correct in both methods. – Kavi Rama Murthy Jun 29 '19 at 11:55 — Kavi Rama Murthy, Jun 29 '19 at 11:55
Are you sure about the second substitution method? – Mohammad Riazi-Kermani Jun 29 '19 at 12:10 — Mohammad Riazi-Kermani, Jun 29 '19 at 12:10

mechanodroid · Accepted Answer · 2019-06-29T12:08:30.473

With the first substitution you get:

$$I = \iint_D(4y- x^2-y^2)\,dxdy=\int_0^\pi \int_0^{4\sin \theta} (4r\sin\theta - r^2)\cdot r\,drd\theta = 8\pi$$

The second substitution should be $x = r\cos\theta, y = 2+r\sin\theta$. The Jacobian is again $r$, and the limits are $r \in [0,2], \theta \in [0,2\pi]$. We get

$$I = \iint_D (-x^2-(y-2)^2+4) \,dxdy = \int_0^{2\pi}\int_0^2 (-r^2+4)\cdot r\,drd\theta = 8\pi$$ which is the same.

To evaluate double integral using Polar coordinates

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