# Integral of surface

$$\iint_\limits{S}\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}dS$$ where $$S: \ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$

I tried to solve it : $$\iint_{S^+}P(x,y,z)\ dydz = \iint_{S}P\cos\alpha dS=\iiint_V\frac{\partial P}{\partial x}dxdydz,\ \; \vec{n}=(cos\alpha, cos\beta, cos\gamma)$$

And i get: $$cosa = \frac{x}{\sqrt{ \frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}}$$ After computing $P$ and $\frac{\partial P}{\partial x}$ $$\iiint_\limits V \frac{2}{a^2} - a^2\frac{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}{x^2}dxdydz$$ in spherical system: $$\int_0^1 \int_0^\pi \int_0^{2\pi} \frac{1}{a^2} - \frac{(\cot{\theta}^2 \sec{\phi}^2)}{c^2} - \frac{\tan{\phi}^2}{b^2}d\phi d\theta dr$$ The last integral does not converge. What's next?

• Hint: $$\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}} = \vec{x} \cdot \hat{n} \quad\text{ where }\quad \begin{cases} \vec{x} &= (x,y,z)\\ \hat{n} &= \frac{1}{\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}}\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right) \end{cases}$$ – achille hui May 18 '15 at 17:52
• @achillehui thanks, got it:) – Simankov May 18 '15 at 17:57