UPDATE: My professor tried this problem and got the same answer as me. The answer they had in the system seems to have been incorrect and my answer was correct. I’m getting credit for the question now. Thank you to those who looked over my work and helped me!
The question says: "Integrate the function $H(x,y,z)=6x^2\sqrt{17-8z}$ over the parabolic dome $z=2-2x^2-2y^2,\,z\geq0$."
Here's my work:
$\mathbf{r}(r,\theta)=r\cos\theta\,\mathbf{i}+r\sin\theta\,\mathbf{j}+(2-2r^2)\,\mathbf{k}$
$\mathbf{r}_r\times\mathbf{r}_\theta=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta &\sin\theta & -4r \\ -r\sin\theta & r\cos\theta & 0\end{array}\right|=\mathbin{\color{red}+}4r^2\cos\theta\,\mathbf{i}\mathbin{\color{red}{+}}4r^2\sin\theta\,\mathbf{j}+r\mathbf{k}$
(EDIT: I believe the $\mathbf{j}$-component should be positive here and not negative like I had it before. Either way it shouldn't have made a difference in the answer since $||\mathbf{r}_r\times\mathbf{r}_\theta||$ is unaffected.)
(EDIT 2: Oops, the $\mathbf{i}$-component should be positive too! Either way this still shouldn’t have changed the answer)
$\begin{align} ||\mathbf{r}_r\times\mathbf{r}_\theta||&=\sqrt{16r^4(\cos^2\theta+\sin^2\theta)+r^2}\\ &=\sqrt{16r^4+r^2}\\ &=\sqrt{r^2(16r^2+1)}\\ &=r\sqrt{16r^2+1} \end{align}$
$\begin{align} H\left(\mathbf{r}(r,\theta)\right)&=6r^2\cos^2\theta\sqrt{17-8(2-2r^2)}\\ &=6r^2\cos^2\theta\sqrt{17-16+16r^2}\\ &=6r^2\cos^2\theta\sqrt{16r^2+1} \end{align}$
$\begin{align} \iint\limits_S H(x,y,z)\,d\sigma&=\iint\limits_R H(\mathbf{r}(r,\theta))||\mathbf{r}_r\times\mathbf{r}_\theta||\,dr\,d\theta\\ &=\int\limits_0^{2\pi}\int\limits_0^1 \left[6r^2\cos^2\theta\sqrt{16r^2+1}\left(r\sqrt{16r^2+1}\right)\right]\,dr\,d\theta\\ &=\int\limits_0^{2\pi}\int\limits_0^1 6r^3\cos^2\theta(16r^2+1)\,dr\,d\theta\\ &=\int\limits_0^{2\pi}\int\limits_0^1 (96r^5+6r^3)\cos^2\theta\,dr\,d\theta\\ &=\int\limits_0^{2\pi} \left(16r^6+\frac{3}{2}r^4\right)\Bigg|_{r=0}^{r=1}\,\cos^2\theta \,d\theta\\ &=\int\limits_0^{2\pi} \frac{35}{2}\cos^2\theta\,d\theta\\ &=\int\limits_0^{2\pi} \frac{35}{2}\left(\frac{1+\cos2\theta}{2}\right)\,d\theta\\ &=\frac{35}{4}\int\limits_0^{2\pi}(1+\cos2\theta)\,d\theta\\ &=\frac{35}{4}\left(\theta+\frac{1}{2}\sin 2\theta\right)\Bigg|_0^{2\pi}\\ &=\boxed{\frac{35}{2}\pi} \end{align}$
The answer was supposed to be $35\pi$. Where did I go wrong?