S is the graph $z=25-(x^2+y^2)$ over the disk $x^2+y^2\leq 9$ and $\varphi = z^2dx\wedge dy$. Find $\int_S \varphi$.
According to the book the answer is $3843\pi$, but the answer I got is different.
What I did: First define $k(x,y)=(x,y,25-(x^2+y^2))$ then $k_x=(1,0,-2x)$ and $k_y=(0,1,-2y)$.
Now, I have that $\varphi (k_x,k_y)=(25-(x^2+y^2))^2dx\wedge dy (k_x,k_y)=(25-(x^2+y^2))^2$.
I ought to find $\int_S\varphi$ which will be given by $$\int_S\varphi=\int_0^3\int_0^{\sqrt{9-y^2}}625-50(x^2+y^2)+(x^2+y^2)^2dxdy$$.
To make it easier, I changed to polar coordinates: $x=r\cos\phi,y=r\sin\psi$. Then I have
$$\int_S\varphi = \int_0^{\color{red}{2\pi}}\int_0^3625r-50r^3+r^5drd\phi\\=\int_0^{\color{red}{2\pi}}\left(\frac{625}{2}(3)^2-\frac{50}{4}(3)^4+\frac{1}{6}(3)^6\right)d\phi\\=\int_0^{\color{red}{2\pi}}\left(\frac{5625}{2}-\frac{4050}{4}+\frac{726}{6}\right)d\varphi\\=\int_0^{\color{red}{2\pi}} 1921\; d\phi\\=3842\pi.$$
Looks like I have missed something, maybe my change to polar coordinates is wrong or there's another mistake that I cannot see.
Update:Fixed the limits of integration
$3843\pi$ is correct. I would recommend doing the integration slightly more economically. We want $$2\pi \int_0^3 (25-r^2)^2 r\,dr = \pi \int_{16}^{25} u^2\, du,$$ having made the substitution $u=25-r^2$. :)