Evaluation of surface integral $\iint F ds$; $F(x,y,z)=(x,y,z)$

Question

Let $F(x,y,z)=(x,y,z)$. $\\[10pt]$ Evaluate: $\iint_{S} F ds$; where $S$ is the upper hemisphere of radius $3$, centered at the origin.

I defined $\phi(u,v)=(\sqrt{9-v^2}\cos(u),\sqrt{9-v^2}\sin(u),v); 0\leq u \leq 2\pi; 0\leq v \leq 3$. Then, I arrived at $\phi_u \times \phi_v=\phi(u,v)$, so: $$\iint_{S} F ds=\int_{0}^{2\pi}\int_{0}^{3}||\phi(u,v)||^2 dudv=\iint_{S} F ds=\int_{0}^{2\pi}\int_{0}^{3}9 dudv=54\pi$$ Is this approach correct?

Answer 1

Just to give something different…

If we close off the surface by adding the disk of radius 3 in the xy plane, we can apply the divergence theorem.

$\iint F dS + \iint F dA = \iiint \nabla \cdot F dV$

The integral over the disk equals zero, which can be deduced from the symmetry of F.

The surface integral over hemisphere is 3 times the volume of the hemisphere.

Answer 2

Yes, your calculation gives the correct answer. A minor point on notation: usually we use $\iint_S \vec{F}\cdot d\vec{s}$ to mean integrating a vector field on a surface. Writing $\iint_S Fds$ can be slightly confusing when both scalar and vector fields are involved.

Here is a simpler approach which does not require parametrization.

Since $S$ is the upper hemisphere of radius $3$ centered at the origin, we can easily identify the outward pointing unit normal vector at $(x,y,z)$ to be $(x,y,z)/(x^2+y^2+z^2)=\frac{1}{3}(x,y,z)=\frac{1}{3}\vec{F}(x,y,z)$. Therefore the integral becomes

$$\dfrac{1}{3}\iint_S (\vec F\cdot \vec F) ds=\dfrac{1}{3}\iint_S ||\vec{F}||^2 ds=\dfrac{9}{3}\iint_S ds=3\text{Area}(S)=3\cdot 2\cdot \pi \cdot 3^2=54\pi$$

This avoids the calculation of cross product.

Answer 3

The end results $54\pi$ is correct but the chain of equations by which it was derived is very confusing.

This answer shows various methods to calculate a surface integral $$ \iint_S\mathbf{F\cdot n}\,dS\,. $$ I prefer this notation to make clear that we are dotting the vector $\mathbf{F}$ with the outer unit normal vector to $S\,.$ Using your parametrization $$ \phi(u,v)=\begin{pmatrix} \sqrt{r^2-v^2}\cos u\\ \sqrt{r^2-v^2}\sin u\\ v \end{pmatrix} $$ the unnormalized normal vector to $S$ is $$ \phi_u\times\phi_v=\begin{pmatrix}-\sqrt{r^2-v^2}\sin u\\ \sqrt{r^2-v^2}\cos u\\ 0\end{pmatrix}\times\begin{pmatrix}\frac{-v}{\sqrt{r^2-v^2}}\cos u\\ \frac{-v}{ \sqrt{r^2-v^2}}\sin u\\ 1\end{pmatrix}=\begin{pmatrix}\sqrt{r^2-v^2}\cos u\\\sqrt{r^2-v^2}\sin u\\v\end{pmatrix}\,. $$

• In this symmetric case this cross product equals $\phi(u,v)$ and $\mathbf{F}\,.$

In the linked answer I also elaborated at length why the surface integral becomes \begin{align} \iint_S\mathbf{F\cdot n}\,dS=\int_0^{2\pi}\int_0^rr^2-v^2+v^2\,dv\,du =2\pi r^3\,. \end{align} In short: we use the unnormalized $\phi_u\times\phi_v$ to make this integral immune against reparametrization.

One could get the results much quicker by noticing that $\mathbf{F\cdot n}$ is constant on $S$ and equals $$ \frac{\|\mathbf F\|^2}{\|\mathbf F\|}=\|\mathbf F\|=r\,. $$ Then the integral is $r$ times half of the surface of the sphere $r\cdot 2\pi r^2\,.$