I have to solve the following problem:
A cylinder of radius R in $\mathbb{R}^3$ with axis along the $z$-axis of a Cartesian coordinate system $(x,y,z)$ can be parametrised as $\textbf{x}(\theta,z) = (R\cos\theta, R\sin\theta, z)$. Evaluate the the surface integral of the vector field $\textbf{A} = (xz,0,yz)$ over the portion of the surface that lies between $z=0$ and $z=R$.
My Approach
- I found the outward-pointing unit normal to the cylinder at the point $\textbf{x}(\theta,z)$. $$ \frac{\partial \textbf{x}}{\partial \theta} \times \frac{\partial \textbf{x}}{\partial z} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k}\;\\ -R\sin\theta & R\cos\theta & 0\\ 0 & 0 & 1 \end{vmatrix} = R\cos\theta\,\textbf{i} + R\sin\theta\,\textbf{j} $$ so $$ \biggl\lVert\frac{\partial \textbf{x}}{\partial \theta} \times \frac{\partial \textbf{x}}{\partial z} \biggr\rVert = \sqrt{(R\cos\theta)^2 + (R\sin\theta)^2} = \sqrt{\!R^2} = R. $$
Thus our outward-pointing unit normal is $$ \textbf{n} = \frac{\frac{\partial \textbf{x}}{\partial \theta} \times \frac{\partial \textbf{x}}{\partial z}}{\biggl\lVert \frac{\partial \textbf{x}}{\partial \theta} \times \frac{\partial \textbf{x}}{\partial z}\biggr\rVert} = \begin{bmatrix} \cos\theta \\ \sin\theta\\ 0\\ \end{bmatrix}. $$
- I rewrote $\textbf{A}$ as $\textbf{A} = (Rz\cos\theta, 0, Rz\sin\theta)$.
- I Rewrote and solved the integral: \begin{align} \iint_{S}^{} \textbf{A}\,d\textbf{S} &= \int_{0}^{R} \int_{0}^{2\pi} \textbf{A} \cdot \biggl( \frac{\partial \textbf{x}}{\partial \theta} \times \frac{\partial \textbf{x}}{\partial z} \biggr)\, d\theta\, dz \\ &= \int_{0}^{R} \int_{0}^{2\pi} (Rz\cos\theta, 0, Rz\sin\theta) \cdot (R\cos\theta, R\sin\theta, 0)\, d\theta\, dz \\ &= \int_{0}^{R} \int_{0}^{2\pi} R^2z\,(\cos\theta)^2\, d\theta\, dz \\ &= \int_{0}^{R} R^2\pi z\, dz \\ &= \frac{R^4\pi}{2}. \end{align}
Is anything wrong with my solution? Am I missing something? Any input would be deeply appreciated.