Let $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$, $F(x,y,z)=(x,0,-x)$ and $n(x,y,z)$ be the unit normal vector of $S$ such that $n(0,0,-1)=(0,0,-1)$.
I want to evaluate $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS$.
My Attempt
Let $f(x,y,z)=x^2+y^2+z^2-1$. Then $n$ can be calculated by $n=\frac{\nabla f}{|\nabla f|}=(x,y,z)$. This satisfies the condition stated in the problem.
Therefore we have $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS=\iint_{S}(x^2-zx)dS$. Now we need to calculate this surface integral, but I'm encountering issues.
According to this website, I have two options. One option is to find an orthogonal projection of $S$. The other option is to find a parameterization of $S$. However, I couldn't do either of them. Is there a simple expression for them? Any help is appreciated.
Best Answer
I will present three ways of tackling this problem.
$\textbf{Option 1}$: Directly
Parametrizing spherical coordinates as usual we can find the bounds by examining the plane equation
$$x+y+z = 1 \implies \sin\phi\cos\theta+\sin\phi\sin\theta+\cos\phi = 1$$
which after a little manipulation becomes
$$\cos\theta+\sin\theta = \frac{1-\cos\phi}{\sin\phi} = \tan\left(\frac{\phi}{2}\right)$$
It's obvious from looking at the sphere with the planar cap cutoff that doing the $\theta$ integral first would require two integrals (one each for the upper and lower parts), but $\phi$ first would only require one. Writing the integral gets us
$$\int_0^{2\pi}\int_{2\cot^{-1}\left(\frac{1}{\sin\theta+\cos\theta}\right)}^\pi \sin^3\phi\cos^2\theta-\sin^2\phi\cos\phi\cos\theta\:d\phi d\theta$$
$\textbf{Option 2}$: Rotate then directly
Once we have the scalar surface integral $\iint_S x^2-xz\:dS$ we can consider a change of variables to rotate the plane so it is completely horizontal from the top. The closest distance between the plane and the origin is $\frac{1}{\sqrt{3}}$ so hopefully that motivates the following rotational coordinate change:
$$\begin{cases}u = \frac{x-y}{\sqrt{2}} \\ v = \frac{x+y-2z}{\sqrt{6}} \\ w = \frac{x+y+z}{\sqrt{3}}\\ \end{cases}$$
This change of variables is a pure rotation so it has Jacobian $1$. This was obtained by choosing a direction for the new '$z$' and '$x$' ($w$ and $u$, respectively) then taking their cross product to find the third orthogonal vector (then including the factor to make them unit vectors), which means
$$u^2+v^2+w^2 = x^2+y^2+z^2$$
Since this is a rotation matrix, inverting the system of equations is as easy as taking the transpose:
$$\begin{cases}x = \frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ y = -\frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ z = -\frac{2v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ \end{cases}$$
which gives us a new surface integral on the same sphere, only rotated
$$\iint_S \frac{u^2+v^2}{2}+\frac{2uv}{\sqrt{3}}+\frac{uw}{\sqrt{6}}+\frac{vw}{\sqrt{2}}\:dS$$
The nice thing about this is that with this rotated sphere cutoff at $w=\frac{1}{\sqrt{3}}$, we can now exploit symmetry. $u$ and $v$ are both odd functions, so any term with just an odd power of either will vanish, leaving us with
$$\iint_S \frac{u^2+v^2}{2}\:dS = \int_0^{2\pi}\int_{\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)}^\pi \frac{1}{2}\sin^3\phi\:d\phi d\theta$$
which easy enough to evaluate since the integrand will end up being only cosines.
$\textbf{Option 3}:$ Divergence theorem
We close off the surface with a plane and consider the equation
$$\iint_{\text{sphere}}F\cdot dS + \iint_{\text{plane}}F\cdot dS = \iiint_V \nabla \cdot F dV$$ $$ = \int_0^{2\pi}\int_{\frac{1}{\sqrt{3}}}^1 \int_{\cos^{-1}\left(\frac{1}{\sqrt{3}\rho}\right)}^\pi\rho^2\sin\phi\:d\phi d\rho d\theta + \frac{4}{3}\pi\left(\frac{1}{\sqrt{3}}\right)^3$$
Since $\nabla\cdot F = 1$, we only want the volume, thus we have implicitly taken advantage of the rotation from option 2 to set up our bounds. The remaining surface integral can be calculated as follows
$$\iint_{\text{plane}}F\cdot dS = \iint_{\text{plane}}(x,0,-x)\cdot\frac{(1,1,1)}{\sqrt{3}}dS = 0$$
therefore the answer will only be the volume integral.