$$\iint_{S^+}x^3dydz $$where $S$ bottom part of $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
I tried to solve it by formula : $$\iint_{S^+}P(x,y,z)dydz = \iint_{S}P\cos\alpha dS,\ \; \vec{n}=(cos\alpha, cos\beta, cos\gamma)$$
and i get $$\iint_S \frac{x^4}{a^2\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{y^2}{c^4}}}dS$$ then i have formula $$\iint_S f(x,y,z)dS = \iint f(x(y,z),y,z)\sqrt{1+(\frac{\partial x}{\partial y})^2+(\frac{\partial x}{\partial z})^2 }$$ but i get stuck with next integral. I am on right way?
$\def\F{{\bf F}}\def\S{{\bf S}}\def\k{{\bf k}}$
You're integrating the vector field $\F = (x^3,0,0)$ over $S$, the bottom half of the oriented ellipsoid. If we let $D$ be the solid region enclosed by $S$ and a flat "top" lying in the plane $z = 0$, we may apply the divergence theorem to get $$ \iint_S \F \cdot d\S = \iiint_D 3x^2\;dV - \iint_{\text{top}\kern-4pt} \F \cdot \k \;dS$$ Now $\F\cdot \k =0$, so the integral over the top vanishes. In the triple integral, we can transform the ellipsoid to the unit ball by scaling $x,y,z$ by $a,b,c$ to get $$\iiint_D 3x^2\;dV = 3a^3bc \iiint\limits_{\text{half ball}} x^2\;dV = (3a^3bc)\; {1\over6}\iiint\limits_{\text{unit ball}} (x^2+y^2+z^2)\;dV$$ where the last integral is obtained by symmetry: $$\iiint\limits_{\text{half ball}} x^2\;dV = {1\over2}\iiint\limits_{\text{unit ball}} x^2\;dV = {1\over2} \iiint\limits_{\text{unit ball}} y^2\;dV = {1\over2} \iiint\limits_{\text{unit ball}} z^2\;dV\,.$$ Now $$\iiint\limits_{\text{unit ball}} (x^2+y^2+z^2)\;dV = \iint\limits_{\text{unit sphere}} \left\lbrace\int_0^1\rho^2 \cdot \rho^2 \;d\rho\right\rbrace\;dS = {1 \over 5}\cdot 4\pi\,.$$
Putting it all together we get $$ \iint_S x^3 \; dy \wedge dz = \iint_S \F \cdot d\S = {2 \over 5} \pi a^3bc\,.$$