# Evaluate the double integral by changing to polar coordinates

I experience some difficulty with converting to polar coordinates in integrals.

So the question I'm struggling with is

Evaluate the double integral $$\int\int x^{6}y\, dA$$ where $D$ is the top half of the disc with center the origin and radius $4$, by changing to polar coordinates.

I'm not sure about solving this one.

I think that I have to use the $r$ going from $0$ to $4$ and since it's the top half I have to go from $0$ to $\pi$. My question is about how to go from $x^{6}y$ to the polar form.

Hopefully someone can help me with this one.

$x = r\cos \theta$ and $y = r\sin \theta$
So $x^6y = r^6\cos ^6 \theta * r\sin \theta = r^7 \cos^6 \theta * \sin \theta$
So the integral is $\int_{0}^{\pi}\int_{0}^{4} r^7 \cos^6 \theta * \sin \theta r\,dr\,d\theta$
Let $u = \cos \theta$ then $du = -\sin \theta \ d\theta$
Then $\int_{0}^{\pi}\int_{0}^{4} r^8 \cos^6 \theta * \sin \theta \,dr\,d\theta$ = -$\int_{1}^{-1}\int_{0}^{4} r^8 u^6* \,dr\,du$ = -($\frac{4^9}{9}*\frac{-2}{7}) = \frac{2^{19}}{63}$