Calculate $\int_M x_1x_3dS_2$ with $M:=\{ x\in \mathbb{R}^3 x_1^2+x_2^2+x_3^2=4, x_1 \leq 0 x_2\geq 0 x_3 \geq 0 \}$

Question

So I have this problem:

I have to calculate

$\int_M x_1x_3dS_2$

With $M:=\{ x\in \mathbb{R}^3 x_1^2+x_2^2+x_3^2=4, x_1 \leq 0 x_2\geq 0 x_3 \geq 0 \}$

Our Professor gave us this solutions that I cannot understand

$\phi:[0, \pi/2)x[\pi/2,\pi) \rightarrow \mathbb{R}^3$ $(\alpha, \beta)\rightarrow 2(\sin(\alpha)\cos(\beta),\sin(\alpha)\sin(\beta),\cos(\alpha))$

and then he has written $\int_{M}x_2dS_2=\int_{0}^{\pi/2}\int_{\pi/2}^{\pi}4\sin(\alpha)\cos(\beta)\cos(\alpha)4\sin(\alpha)d\beta d\alpha=-16\int_{0}^{\pi/2} \sin^2(\alpha)\cos(\alpha)d\alpha=-16/3$ So what I don't understand is:

1. Why and how did he create the function $\phi$?
2. Why did he write $\int_{M}x_2dS_2$ instead of $\int_M x_1x_3dS_2$?
3. I don't understand which formula did he used to write $4\sin(\alpha)\cos(\beta)\cos(\alpha)4\sin(\alpha)$ inside the integral.

Answer 1

1. The standard sphere parametrization of radius $R$ centered at the origin is $$ \Phi(\alpha, \beta) = (R \sin(\alpha)\cos(\beta), R \sin(\alpha)\sin(\beta),R \cos(\alpha)),\quad (\alpha, \beta) \in [0, \pi] \times [0, 2 \pi). \quad (1) $$ The equation $$ x_1^2+x_2^2+x_3^2=4 $$ defines the sphere of radius $2$ centered at the origin. The inequalities $$ x_1 \leq 0, x_2\geq 0, x_3 \geq 0 $$ define a part of this sphere named $M$. So, we can use (1) to parametrize our set $M$, but we need to take only $(\alpha, \beta)$ correspond to the set $M$. We have $$ x_1 = \sin(\alpha)\cos(\beta) \leq 0, x_2 = \sin(\alpha)\sin(\beta)\geq 0, x_3 \cos(\alpha) \geq 0, (\alpha, \beta) \in [0, \pi] \times [0, 2 \pi) $$ The solution of this system is $(\alpha, \beta) \in [0,\pi/2) \times [\pi/2,\pi)$. So the parametrization of $M$ is $$ \phi(\alpha, \beta) = (2 \sin(\alpha)\cos(\beta), 2 \sin(\alpha)\sin(\beta), 2\cos(\alpha)),\quad (\alpha, \beta) \in [0, \pi /2) \times [\pi/2,\pi). \quad (2) $$

2. I think it was a typo.

3. Let us calculate the first fundamental form of $M$. $$ \phi_\alpha(\alpha,\beta) = (2 \cos (\alpha ) \cos (\beta ),2 \cos (\alpha ) \sin (\beta ),-2 \sin (\alpha )), \\ \phi_\beta(\alpha,\beta) = (-2 \sin (\alpha ) \sin (\beta ),2 \sin (\alpha ) \cos (\beta ),0). $$ So, $$ E = \langle\phi_\alpha\mid\phi_\alpha\rangle = 4, \\ F = \langle\phi_\alpha\mid\phi_\beta\rangle = 0, \\ G = \langle\phi_\beta\mid\phi_\beta\rangle = 4 \sin ^2(\alpha ), $$ and $$ \sqrt{EG - F^2} = 4 \sqrt{\sin ^2(\alpha )}.$$ Because $\alpha \in [0, \pi /2)$ we have $$ \sqrt{EG - F^2} = 4 \sin(\alpha ) $$ Now we calculate the surface element $dS_2$ $$ dS_2 = \sqrt{EG - F^2} \, d\alpha\, d\beta = 4 \sin(\alpha ) \, d\alpha\, d\beta $$ and $x_1 x_3$ using the obtained parametrization $$ x_1 x_3 = 4 \sin(\alpha)\cos(\beta) \cos(\alpha). $$ So, we have $$ \int\limits_M x_1 x_3 \,d S_2 = \iint\limits_{(\alpha, \beta) \in [0,\pi/2) \times [\pi/2,\pi)} 4 \sin(\alpha)\cos(\beta) \cos(\alpha) 4 \sin(\alpha ) \, d\alpha\, d\beta = \\ = \int\limits_{0}^{\pi/2} d\alpha \int\limits_{\pi/2}^\pi 4 \sin(\alpha)\cos(\beta) \cos(\alpha) 4 \sin(\alpha ) \, d\beta. $$

Answer 2

Your professor used spherical coordinates to parametrize the patch on the sphere of radius 2 [https://en.wikipedia.org/wiki/Spherical_coordinate_system]: $x_1= 2\sin{\alpha}\cos{\beta}$, $x_2 = 2\sin{\alpha}\sin{\beta}$, $x_3 = 2\cos{\alpha}$.

The coordinate domain $[0, \pi/2] \times [\pi/2, \pi]$ was chosen to match the restrictions $x_1 \leq 0$, $x_2\geq 0$, $x_3 \geq 0$.

The part about $\int_M x_2 \, dS_2$ seems to be a typo, but in what follows the correct $\int_M x_1x_3\, dS_2$ is computed.

When computing the integral in the $(\alpha, \beta)$-parameter space: $x_1x_3 = 2\sin{\alpha}\cos{\beta} \cdot 2\cos{\alpha}$ and $dS_2 = 2^2\sin{\alpha}\;d\alpha\; d\beta$. [See the "surface element" link from the above page, or the relevant section of your textbook.]

Hence $x_1x_3 \; dS_2$ is transformed to $4\sin{\alpha}\cos{\beta}\cos{\alpha} \cdot 4\sin{\alpha} \; d\alpha\; d\beta$.