How are surface integrals converted from dS to dA?

Question

In class, we learned that:

$$ \iint \vec{F} \cdot d \vec{S} = \iint \vec{F} \cdot \hat{n} \|r_{u} \times r_{v}\| dA = \iint \vec{F} \cdot \hat{n} \|r_{u} \times r_{v}\| dudv, $$ where the normal unit vector is equal to $\frac{r_{u} \times r_{v}}{\|r_{u} \times r_{v}\|}$.

Moreover in addition to this, the bounds of the double integral depend on the parametrization $r_{u}$ and $r_{v}$, or any other vector that is normal to the surface we are integrating over (for example, the three standard unit vectors, or a gradient vector).

However, in this problem:

I'm really confused by the second part, where they had to take the surface integral of the bottom disk of the top hemisphere.

It seems like they bypassed the entire parametrization $dS = \|r_{u} \times r_{v}\|dA$.

How do I make sense of this? Did something happen that was not shown?

Answer 1

This is kind of a subtle question. So I want to answer the question itself systematically and hopefully in turn answer any other questions you may have.

First, note the problem is asking us to find the surface integral $\iint_{S}\mathbf F\cdot\mathbf n \space dS$ (otherwise known as flux) of $\mathbf{F}$ across the surface $S$, where $S$ is the top half of the unit sphere $x^2+y^2+z^2=1$ oriented upward. It's important to understand that this is not a closed surface i.e. the surface $S$ only consists of points on the upper hemisphere when $z>0$ and not the equatorial disk $x^2+y^2\leq1, z=0$.

What is meant by orienting $S$ upward? Orienting a surface $S$ upwards means that the normal vector induced by a chosen parametrization $\mathbf r(u,v)$ of the surface $S$ :$$\textbf{n}=\frac{\mathbf r_u \times \mathbf r_v}{|\mathbf r_u \times \mathbf r_v|}$$ must have positive $z$ - coordinate for all parameter values. So if we parametrize our unit, upper hemisphere using spherical coordinates, we obtain $$\mathbf r(u,v)=\left(\sin(u)\cos(v)\right)\mathbf i + \left(\sin(u)\sin(v)\right)\mathbf j+ \left(\cos(u)\right)\mathbf k$$ where $0\leq u < \pi/2$ and $ 0\leq v\leq2\pi$ (note I did not include $\pi/2$ for my $u$-parameter so that the $z$-coordinate and is nonzero and positive). As an exercise for yourself if you want, one can then obtain that $$\mathbf r_u \times\mathbf r_v = \left(\sin^2(u)\cos(v)\right)\mathbf i + \left(\sin^2(u)\sin(v)\right)\mathbf j+ \left(\sin(u)\cos(u)\right)\mathbf k $$ and $$|\mathbf r_u \times \mathbf r_v|=\sin(u)$$ so that $$\mathbf n=\frac{\mathbf r_u \times \mathbf r_v}{|\mathbf r_u \times \mathbf r_v|}=\left(\sin(u)\cos(v)\right)\mathbf i + \left(\sin(u)\sin(v)\right)\mathbf j+ \left(\cos(u)\right)\mathbf k=\mathbf r(u,v)$$ which has positive $z$-coordinate (since $\cos(u)>0$ when $0\leq u < \pi/2$) for all parameter values. So the unit normal vector $\mathbf n$ induced by the parametrization we gave for the upper hemisphere orients the surface upward.

Now, you should picture how this looks. The normal vector $\mathbf{n}$ at any point on the upper hemisphere will point in the same direction as the position vector $\mathbf r(u,v)$ at that point, and emanate from the surface of the sphere. Note these normal vectors on the upper hemisphere coincide with the normal vectors on the upper hemisphere of the closed surface $S_2$ if $S_2$ is oriented outward as required by the Divergence Theorem.

But most importantly, if $S_2$ (again remember $S_2$ is closed; we've capped off the upper hemisphere $S$ when $z>0$ with the disk $x^2+y^2\leq1$ when $z=0$) is oriented outward, the bottom cap of the surface i.e. the disk $x^2+y^2\leq1, z=0$, has normal vectors emanating straight downwards in the $-\mathbf k$ direction. Hence, the outward orientation on $S_2$ induces a downward orientation on the standalone disk $x^2+y^2\leq1, z=0$.

So, using the Divergence Theorem across the closed surface $S_2$ with outward orientation calculates the flux across our surface of interest with the same orientation, the upper hemisphere, AND the flux across the bottom disk $x^2+y^2\leq1, z=0$ orientated outward when considered part of the closed surface or oriented downward when considered alone. Since we're only interested in the flux across the upper hemisphere, we calculate the flux across the disk $x^2+y^2\leq1, z=0$ when the disk is oriented downward, and then subtract this from the flux across $S_2$ found using the Divergence Theorem.

Now, let us first calculate the flux through the surface $S_1$ using the definition of flux: $$\Omega_{\mathbf F,S_1}=\iint_{S_1} \mathbf F\cdot \mathbf n \space dS = \iint_{S_1}\mathbf F \cdot d\mathbf S$$ where $\mathbf n$ is defined in terms of $\mathbf r$ as the unit vector above, $dS = |\mathbf r_u \times \mathbf r_v |\space dA$, and $d\mathbf S = (\mathbf r_u \times \mathbf r_v)\space dA$.

We know ahead of time from the reasoning above that $\mathbf n$ for the standalone disk $x^2+y^2\leq1, z=0$ will be oriented downward and equal to $-\mathbf k$, but we can also parametrize it as $\mathbf r(r,\theta)=\left(r\sin\theta\right)\mathbf i + \left(r\cos\theta\right)\mathbf j + 0\mathbf k$ where $0\leq r \leq 1$ and $0\leq \theta \leq 2\pi$ to induce a negative orientation on the disk and utilize the formulas alluded to by OP. Thus $$\mathbf r_r=\left(\sin\theta\right)\mathbf i + \left(\cos\theta\right)\mathbf j + 0\mathbf k$$ $$\mathbf r_{\theta}=\left(r\cos\theta\right)\mathbf i + \left(-r\sin\theta\right)\mathbf j + 0\mathbf k$$ so $$\mathbf r_r \times \mathbf r_{\theta} = \left(0\right)\mathbf i + \left(0\right)\mathbf j + \left(-r\sin^2\theta - r\cos^2\theta\right) \mathbf k=\left(0\right)\mathbf i + \left(0\right)\mathbf j + \left(-r\right) \mathbf k$$ and finally $$\mathbf n = \frac{\mathbf r_r \times \mathbf r_{\theta}}{|\mathbf r_r \times \mathbf r_{\theta}|}=\frac{\left(0\right)\mathbf i + \left(0\right)\mathbf j + \left(-r\right) \mathbf k}{r}=\left(0\right)\mathbf i + \left(0\right)\mathbf j + \left(-1\right) \mathbf k$$ which is the downward orientation we were looking for. Putting this all together then, the flux through the surface $S_1$ is $$\Omega_{\mathbf F,S_1} = \iint_{S_1}\mathbf F \cdot d\mathbf S=\iint _{D_1}\mathbf F \cdot \left(\mathbf r_r \times \mathbf r_{\theta}\right)\space dA$$

$$=\int_0^{2\pi}\int_0^1\mathbf F(\mathbf r(r,\theta))\cdot \left(0 \mathbf i + 0 \mathbf j + (-r) \mathbf k\right)\space dA$$ $$=\int_0^{2\pi}\int_0^1\left(r\cos\theta\right)^2\cdot (-r)\space dr \space d\theta=-\left(\int_0^{2\pi}\cos^2\theta \space d\theta\right)\cdot\left(\int_0^1r^3 \space dr\right)$$ $$=-\frac{1}{4}\int_0^{2\pi}\left(\frac{1+\cos(2\theta)}{2}\right)\space d\theta=-\frac{\pi}{4}$$

As in your provided work, we could use the formula $\Omega_{\mathbf F,S_1}=\iint_{S_1} \mathbf F\cdot \mathbf n \space dS$ to calculate the flux without any reference to a parametrization (other than knowing $z=0$) for the disk since we know the normal vector ahead of time will be $\mathbf n = -\mathbf k$: $$\Omega_{\mathbf F,S_1}=\iint_{S_1} \mathbf F\cdot \mathbf n \space dS=\iint_{D_1} \mathbf F\cdot \mathbf (-\mathbf k) \space dA=-\iint_{x^2+y^2\leq1}y^2\space dA$$ which we may use polar coordinates to evaluate with $x=r\cos\theta$, $y=r\sin\theta$, and $dA= r\space dr \space d\theta$.

Lastly, to finish out the problem, let's calculate the flux across the closed surface $S_2$ oriented outward using the Divergence Theorem, and then subtract from that result the flux we found across the surface $S_1$ oriented downward: $$\text{div}\space\mathbf F = x^2 + y^2 + z^2\implies$$ $$\iint_{S_2}\mathbf F\cdot \mathbf n \space dS =\iiint_{x^2+y^2+z^2\leq1, z\geq0}\left(x^2+y^2+z^2\right)\space dV$$

In spherical coordinates, the latter integral is equal to $$\int_0^{\pi/2}\int_0^{2\pi}\int_0^{1}\left(\rho^2\right)\cdot\left(\rho^2\sin\phi \space d\rho \space d\theta \space d\phi \right)$$ where $\phi$ ranges from only $0$ to $\pi/2$ since we're only concerned with the solid upper hemisphere where $z\geq 0$. The triple integral evaluates to $\frac{2\pi}{5}$, so that the overall flux through the upper hemisphere without the bottom cap is $$\Omega_{\mathbf F,S}=\Omega_{\mathbf F,S_2}-\Omega_{\mathbf F,S_1}=\frac{2\pi}{5}-\left(-\frac{\pi}{4}\right)=\frac{13\pi}{20}$$

Answer 2

They didn't use a parameterization. They applied the divergence theorem to transform the surface integral into a volume integral, which they then computed using polar coordinates. The divergence theorem says that if $\Omega \subset \mathbb{R}^3$ is a smoothly bounded region, then $$\int_{\Omega}\text{div}(F)\,dx\,dy\,dz = \int_{\partial \Omega}F \cdot n\,dS.$$ They took $\Omega$ to be the top half of the unit ball in $\mathbb{R}^3$, so that $\partial \Omega$ consists of the two pieces $S$ and $S_1$, i.e. $\partial \Omega = S_2$.

Answer 3

It's mostly fair to say that since the disc is in the $xy$-plane, we don't need a transformation or projection, and can treat it as an ordinary two-dimensional double integral.

But it's still true that the disc (and the $xy$-plane) is embedded in 3-space $\mathbb{R}^3$, so the general formula for surface integrals in that context should apply. And maybe technically should be the method, not just switching to a plane integral.

So one way of understanding the integral on the disc in the $xy$-plane is that we have a very simple coordinate system for the whole plane, no matter what the shape of the domain surface: $u=x$, $v=y$. But then $\|r_u \times r_v\| = \|\hat \imath \times \hat \jmath\| = \|\hat k\| = 1$. So the surface integral using the formula you quoted does give the same result as the simple double integral $\iint \vec F \cdot \hat n\, dA$.

With the $\|r_u \times r_v\|$ term out of the picture, we can make a transformation on the double integral using polar coordinates and the usual $dx\, dy = r\, dr\, d\theta$, since in this case it gives easier limits of integration.