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:

  1. 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.

  • 1
    $\begingroup$ You should probably change those $\cos(x)$ and $\sin(y)$ to $\cos(\varphi)$ and $\sin(\theta)$ to avoid confusion. As for the rest, I don't understand what $p$ is. Is it the length of the normal vector? $\endgroup$ – Mark Fantini Dec 22 '14 at 15:52
  • $\begingroup$ @MarkFantini The question does not mention anything else about p. But as per another example, this p must be the length of the perpendicular. $\endgroup$ – Deepabali Roy Dec 22 '14 at 16:39
  • $\begingroup$ Your formula for $p$ looks wrong, it should be $$p = \frac{1}{\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}}$$ or in terms of parametrization $(x,y,z) = (a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta)$, $$p = \frac{1}{\sqrt{ \frac{\cos^2\theta\cos^2\phi}{a^2} + \frac{\cos^2\theta\sin^2\phi}{b^2} + \frac{\sin^2\theta}{c^2} }}$$ $\endgroup$ – achille hui Dec 22 '14 at 16:42
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    $\begingroup$ $p$ is not the length of the normal vector. It is the closest distance between the tangent plane and the origin. The gradient give you a direction of the normal vector and no information about the closest distance to the origin. For the ellipsoid surface at hand, the normal at $(X,Y,Z)$ is in the direction $\left( \frac{2X}{a^2}, \frac{2Y}{b^2}, \frac{2Z}{c^2}\right)$, so the equation of the tangent plane is $$\frac{X}{a^2}(x-X) + \frac{Y}{b^2}(y-Y) + \frac{Z}{c^2}(z - Z) = 0 \quad\iff\quad \frac{xX}{a^2} + \frac{yY}{b^2} + \frac{zZ}{c^2} = 1 $$ $\endgroup$ – achille hui Dec 22 '14 at 17:04
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    $\begingroup$ Given any plane of the form $Ax+By+Cz = 1$, the closest distance between it and the origin is simply given by the formula $\frac{1}{\sqrt{A^2+B^2+C^2}}$. $\endgroup$ – achille hui Dec 22 '14 at 17:05

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