Problem is to find the flow of vector field:
$$\vec F=x^2\widehat i+y^2\widehat j+z^2\widehat k$$
through the outer side of a cylindrical surface $x^2+y^2=4$, bounded by planes $z=0$ and $z=8$, but we are only calculating the flux in the cylinder, not through the top and bottom planes.
I have tried using the normal and parameterise the cylinder and use the expression $$\iint\vec F\cdot\widehat n \:dS$$ but I can't get it right. Do you have any suggestions?
Best Answer
Use cylindrical coordinates to parametrize the cylindrical surface $$ \vec{r}(\theta,z)=\langle 2 \cos \theta, 2\sin \theta,z\rangle, \hspace{2mm} \mbox{ where } \hspace{2mm} 0\leq \theta \leq 2\pi \hspace{2mm} \mbox{ and } \hspace{2mm} 0\leq z \leq 8. $$ So the vector field $\vec{F}$ is given by $$ \vec{F} = \langle 4\cos^2 \theta, 4\sin^2\theta,z^2 \rangle, $$ and the normal vector $\vec{N}$ is $$ \vec{N} = \vec{r}_{\theta} \times \vec{r}_z = \left| \begin{pmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ -2\sin \theta & 2\cos \theta & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \right| = \langle 2\cos\theta, 2\sin\theta,0\rangle, $$ where $0\leq \theta \leq 2\pi$, $0\leq z\leq 8$, and $\widehat{i}, \widehat{j}, \widehat{k}$ are the standard unit vectors. Since we want the normal vector to have unit length, $$ \widehat{n} = \frac{\vec{N}}{||\vec{N}||} = \langle \cos\theta, \sin\theta, 0 \rangle. $$ Thus the flux is $$ \begin{align*} \text{Flux} &= \int\int_S \vec{F}\circ \widehat{n}\: dS \\ &= \int_{0}^{8} \int_{0}^{2\pi} \langle 4\cos^2 \theta, 4\sin^2\theta,z^2 \rangle \circ \langle \cos\theta, \sin\theta, 0 \rangle \: d\theta \: dz \\ &= 8 \int_{0}^{2\pi} 4 (\cos^3 \theta+ \sin^3\theta)\: d\theta = \boxed{0}. \end{align*} $$