For $S: x^2+y^2+z^2=R^2$, calculate $$I=\iint_S x^2 \,dy\,dz$$
I almost know how to do it. We have for $D_{yz}: y^2+z^2\le R^2$ $$I = \iint\limits_{D_{yz}}x^2(y,z)\,dy\,dz = \iint (R^2-y^2-z^2) \,dy\,dz$$ am I correct or should I see the cases $x \ge 0$ and $x < 0$ and say one time $x = \sqrt{R^2-y^2-z^2}$ and another time $x = -\sqrt{R^2-y^2-z^2}$ and calculate two integrals?
I'm quite confused.
$x = R\sin u \cos v, \ \ y= R\sin u \sin v, \ \ z = R\cos u$, then
- $r_u = (R\cos u \cos v, R\cos u \sin v, -R\sin v)$
- $r_v = (-R\sin u \sin v, R\sin u \cos v, 0)$
- $r_u \times r_v = (R^2\sin u \cos v\sin v, R^2\sin u \sin^2 v, R^2\cos u \sin u \cos v - R^2\sin u \sin^2 v)$
$S: x^2+y^2+z^2 = R^2$
$dS = \sqrt {1 + (\frac{dx}{dy})^2 + (\frac{dx}{dz})^2} \ dy \ dz$
$ = \displaystyle \frac{R}{x} \ dy \ dz$
As you rightly said, you need to multiply this by $2$. So the integral should be,
$\displaystyle \int_{-R}^{R} \int_{-\sqrt{R^2 - z^2}}^{\sqrt{R^2 - z^2}} 2R \cdot {\sqrt{R^2-y^2-z^2}} \ dy \ dz$
In spherical coordinates, if we parametrize surface as, $x = R \cos\phi, y = R \sin\theta \sin\phi, z = R \cos\theta \sin\phi$ and we know surface area element in spherical coordinates is $R^2 \sin \phi \ d\phi \ d\theta$, the integral becomes,
$\displaystyle \int_0^{2\pi} \int_0^{\pi} (R \cos\phi)^2 R^2 \sin\phi \ d\phi \ d\theta$