Question is to find the surface integral $$\iint p(x^4 + y^4 + z^4) \, \mathbb dS.$$
The given surface is $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1,$$ and $p$ is the length of the perpendicular from the origin to the tangent plane at $(x, y, z)$.
I have attempted solving the question as follows:
- Parametrizing the surface:
$$ x = (a\cos x)\cos y,$$ $$ y = (b \cos x) \sin y ,$$ $$z = c \sin x $$
I am not able to calculate $p$ correctly. This is an example problem and the value of $p$ mentioned in the book is $$p = \frac{1}{ \sqrt{\frac{\cos^2x + \cos^2y}{ a^2} + \frac{\cos^2x + \cos^2y}{ b^2} + \frac{\sin^2x}{c^2}}}.$$
I know that the normal vector to the tangent plane at a point is given by gradient of the Surface vector. I am getting $$p = 2 \sqrt{2} \sqrt{\frac{\cos^2x + \cos^2y}{ a^2} + \frac{\cos^2x + \cos^2y}{ b^2} + \frac{\sin^2x}{c^2}}.$$
I am not able to understand what I am doing wrong. Please help. Thanks a lot.