Calculating the flux of the vector field + Gauss

Question

Calculate the flux $\int_BF · ν dS$ of the vector field $F : \mathbb{R}^3 \to \mathbb{R}^3, F(x, y, z) := (yz, xz, x^2)^T$ through the northern hemisphere B of radius 1 with 0 as its center, i.e. the surface $B := \{(x, y, z) \in \mathbb{R}^3: x^2 + y^2 + z^2 = 1, z > 0\}$

This is what I got so far:

$z=\sqrt{1-x^2-y^2}$

$vdS=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$

$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}(yz, xz, x^2)*vdS\;dydx$

$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} y\sqrt{1-x^2-y^2},x\sqrt{1-x^2-y^2}, x^2 (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1) \ dy\ dx\\ \int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} x^2+2xy\; dy dx\\ \int_{-1}^1 2x^2\sqrt{1-x^2} dx\\ \frac\pi 2*\frac1 4(\frac\pi 2 * \frac1 4) = \frac \pi 4$

EDIT: I replaced the last two lines, but for some reason the answer doesn't match the one given in this post, which I am sure is correct, so if anyone sees my mistake and can kindly point it out, it would be great! found

Is this correct? And I also want to know how can I calculate using the following steps:

(i) Draw B and find an embedding of B.

(ii) Identify the outer normal vector ν on B.

(iii) Finally, calculate $\int_BF · ν dS.$

What do the first two steps add if I can calculate the flux directly?

I thank anyone who can help!

EDIT.NR2 : If I use Gauss theorem to find the flux is it possible for it to be different?

Answer 1

Using spherical coordinates

$ \hat{n} dS = ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \sin \theta \ d\theta \ d \phi $

$ \vec{F} = ( \sin \theta \cos \theta \sin \phi, \sin \theta \cos \theta \cos \phi, \sin^2 \theta \cos^2 \phi ) $

So

$ \vec{F} \cdot \hat{n} dS = (\sin^2 \theta \cos \theta \cos \phi \sin \phi + \sin^2 \theta \cos \theta \cos \phi \sin \phi + \sin^2 \theta \cos \theta \cos^2 \phi ) \sin \theta \ d\theta \ d\phi$

And this simplifies to

$ \vec{F} \cdot \hat{n} dS = \sin^3 \theta \cos \theta (\sin(2 \phi) + \cos^2(\phi) ) \ d\theta \ d\phi$

Integrating with respect to $\phi$ first, from $\phi = 0 $ to $\phi = 2 \pi$, we get the term $\sin(2 \phi) $ cancelled out, and the term $\cos^2 \phi$ contributing a fator of $ \pi $

i.e.

$\displaystyle \int_B \vec{F} \cdot \hat{n} dS = \pi \int_{\theta = 0}^{\dfrac{\pi}{2}} \sin^3 \theta \cos \theta \ d\theta = \dfrac{\pi}{4}$

Answer 2

Part (i) is probably to familiarize yourself with visualizing surfaces. In the figure, $B$ is the transparent hemisphere (orange) and its embedding in the $(x,y)$ plane is the ruled disk shown in the plane $z=0$ (light blue).

Part (ii) is a necessary step in computing the surface integral because the surface element $d\vec S$ depends on the normal vector. But when it comes to computing the normal to $B$, you can choose either $\vec s_u\times \vec s_v$ or $\vec s_v\times\vec s_u$, where $\vec s(u,v)$ is a vector-valued function that parameterizes $B$. Your choice affects the sign of the surface element and integral. The figure from part (i) can help decide which is the correct choice.

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And yes, you can confirm your result with Gauss' theorem.

Note that the theorem applies to closed surfaces, and $B$ is not closed. So suppose we close it up by connecting to $B$ a simple region, say, the disk $D:=\{(x,y)\in\Bbb R^2\mid x^2+y^2\le1\}$. Now apply the theorem to $B\cup D$:

$$\iint_{B\cup D} \vec F\cdot \vec\nu\,d\vec S = \iiint_{\operatorname{int}(B\cup D)} \underbrace{\nabla\cdot \vec F}_{=0} \, dV = 0$$

which means $\iint_B=-\iint_D$. In the region $D$, we have $\vec F=\langle0,0,x^2\rangle$ and the outer normal is $\vec\nu=\langle0,0,-1\rangle$, so the flux across $D$ is

$$\iint_D \vec F\cdot\vec\nu \, d\vec S = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \langle0,0,x^2\rangle\cdot\langle0,0,-1\rangle\,dy\,dx$$

and multiplying by $-1$ makes this agree with your integral.