I need to calculate the flux through a Hemisphere given implicitly:
$M = \{(X,y,z) \in \mathbb{R}^3 \vert x^2+y^2+z^2 = 1, z \geq 0\}$
Or as I parametrised in Polar coordinates:
$\psi(\varphi, \theta) = \left(\begin{array}{cc}\cos(\varphi)\,\sin(\theta) \\ \sin(\varphi)\,\sin(\theta)\\ \cos(\theta)\end{array}\right), \quad \varphi \in[0,2\,\pi] \quad \theta\in[0,\dfrac{\pi}{2}] $
Through the Vector Field: $F = \begin{array}{c}(x^2\,y \quad y^2\,z \quad z^2\,x)\end{array}$
Of course I know flux can be calculated by:
$\Phi = \int_M \langle F, N\rangle\,\mathrm{dA}$ Where $N$ is the unit normal field of the sphere:
$N = \left(\begin{array}{c}-\cos(\varphi)\,\sin(\theta) \\ -\sin(\varphi)\,\sin(\theta) \\ -\cos(\theta)\end{array}\right)$
also $F$ is a composition: $F \circ \psi(\varphi,\theta)$:
$F = \left(\begin{array}{c}\cos^2(\varphi)\,\sin^3(\theta)\,\sin(\varphi) \\ \sin^2(\varphi) \, \sin^2(\theta) \,\cos(\theta) \\ \cos^2(\theta)\,\cos(\varphi)\,\sin(\varphi)\end{array}\right)$
So all gathered together:
$\begin{align}&\Phi = \displaystyle{\int_0^{\dfrac{\pi}{2}} \int_0^{2\,\pi} \bigg\langle \left(\begin{array}{c}\cos^2(\varphi)\,\sin^3(\theta)\,\sin(\varphi) \\ \sin^2(\varphi) \, \sin^2(\theta) \,\cos(\theta) \\ \cos^2(\theta)\,\cos(\varphi)\,\sin(\varphi)\end{array}\right), \left(\begin{array}{c}-\cos(\varphi)\,\sin(\theta) \\ -\sin(\varphi)\,\sin(\theta) \\ -\cos(\theta)\end{array}\right)\bigg\rangle\,\mathrm{d\varphi\,d\theta}}\\ \\
&\Phi = \displaystyle{\int_0^{\dfrac{\pi}{2}} \int_0^{2\,\pi}-\cos^3(\varphi)\,\sin^4(\theta)\,\sin(\varphi)\,\cos(\varphi)-\sin^3(\varphi)\,\sin^3(\theta)\,\cos(\theta)-\cos^3(\theta)\,\cos(\varphi)\,\sin(\varphi)\,\mathrm{d\varphi\,d\theta}}\end{align}$
What results into an integral that seems pretty hard to deal with. Here I'm hanging
Best Answer
You have mistakes in your working. Check the normal vector. Also the z component of $\vec F(\psi(\varphi, \theta))$ has a mistake.
First, a quick check: $ \ \vec F = (x^2y, y^2z, z^2x)$ and unit normal vector to the sphere is $\hat n = (x, y, z)$. So integral of $\vec F \cdot \hat n = x^3y + y^3z + z^3x$ over surface $S: x^2+y^2+z^2 = 1, z \geq 0$ will be zero due to symmetry about xz and yz planes.
Coming back to your working, if you have parametrized the surface as,
$\psi(\varphi, \theta) = (\cos\varphi\sin\theta, \sin\varphi\sin\theta , \cos\theta), \quad \varphi \in[0,2\,\pi] \quad \theta\in\left[0,\dfrac{\pi}{2}\right]$
Then outward normal vector $\vec N = \psi_\theta \times \psi_\varphi = (\sin^2\theta \cos\varphi, \sin^2\theta \sin\varphi, \sin\theta \cos\theta)$
$\vec F = \left(\cos^2\varphi\sin\varphi\sin^3\theta, \sin^2\varphi \sin^2\theta\cos\theta, \cos\varphi\cos^2\theta\sin\theta\right)$
So, flux through the surface is,
$\displaystyle \int_0^{\pi/2} \int_0^{2\pi} \vec F \cdot \vec N \ d\varphi \ d\theta$
$= \displaystyle \int_0^{\pi/2} \int_0^{2\pi} \left(\cos^3\varphi\sin\varphi\sin^5\theta + \sin^3\varphi \sin^4\theta\cos\theta + \cos\varphi\cos^3\theta\sin^2\theta\right) \ d\varphi \ d\theta$
The inner integral is zero and hence the flux is zero.