Evaluate the surface integral $\iint\limits_S F\cdot ds$, where $F(x,y,z) = (x,y,z^2)$ and $S$ is parametrized by $\phi(u,v)=(2\sin(u), 3\cos(u),v)$ with $0\le u \le 2\pi$ and $0\le v \le 1$.
My Attempt
\begin{align}
\vec{T}_u &= (2\cos(u) , -3\sin (u), 0) \\
\vec{T}_v &= (0,0,1)\\
\vec{T}_{u} \times \vec{T}_v &= \begin{vmatrix}
\vec{i}&\vec{j}&\vec{k} \\ 2\cos u & -3\sin u & 0 \\ 0 & 0& 1
\end{vmatrix} = (-3\sin (u) , 2\cos (u), 0).
\end{align}
Then I use the surface integral for vector fields definition
\begin{align}
\iint\limits_S F \cdot ds &= \iint\limits_{D} F(\phi(u,v)) \cdot (\vec{T}_{u} \times \vec{T}_{v}) dA\\
&=\int_{0}^{2\pi} \int_{0}^{1}(2\sin u , 3\cos u, v)\cdot (-3 \sin u , 2\cos u , 0) \ dvdu \\
&=\int_{0}^{2\pi} \int_{0}^{1}(-6 + 12\cos^{2} u) \ dvdu \\
&= \int_{0}^{2\pi} (-6 + \cos^{2}u) \ du = (\dots ) = 12\pi.
\end{align}
I'm not sure if my reasoning is correct and if I missed some step somewhere because this solution seem a bit to simple in my opinion.
Best Answer
The dot product in the third line wasn't correct. It should be
\begin{align} \iint\limits_S F \cdot ds &= \iint\limits_{D} F(\phi(u,v)) \cdot (\vec{T}_{u} \times \vec{T}_{v}) dA\\ &=\int_{0}^{2\pi} \int_{0}^{1}(2\sin u , 3\cos u, v^2)\cdot (-3 \sin u , 2\cos u , 0) \ dvdu \\ &= \int_0^{2\pi}\int_0^1 6\left (\cos^2u - \sin^2(u)\right )dvdu \\ &= \int_0^{2\pi} 6\cos(2u)du \\ &= 0. \end{align}