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I want to solve the following problem, I want to find

$$ \iint_S x \, \mathrm{d}S $$ where S is the part of the parabolic cylinder that lies inside of the cylinder $z = x^2/2$, and in the first octant of the cylinder $x^2 + y^2 = 1$

I was obviously thinking about switching to cylindrical coordinates, but I have problems setting up the problem and finding the limits.

Could I get some tips / help ? =)

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I'd like to help you, but I don't get the picture. This is a 3D plot of your 2 equations. cl.ly/1v3J1i2v1V392u352b2g Can you describe to me again on which surface you want to find your integral? –  chemeng May 16 '12 at 19:40
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1 Answer 1

up vote 1 down vote accepted

The main challenge here is finding a suitable parametrization. Since the boundary of the surface of integration is defined in terms of a cylinder, it makes sense to try cylindrical coordinates. We have:

$$ (x, y, z) = (\rho\cos\phi, \rho\sin\phi, z) $$

Since $\displaystyle z = \frac{x^2}{2}$, the parabolic cylinder has the following parametrization in cylindrical coordinates:

$$ (x, y, z) = \left(\rho\cos\phi, \rho\sin\phi, \frac{(\rho\cos\phi)^2}{2}\right) $$

And the ranges are:

$$ \rho \in [0, 1], \phi \in [0, \frac{\pi}{2}] $$

Here is a plot of this parametric representation:

plot

From there, you have a straightforward surface integral to solve.

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I thought for a moment that $\rho \in [0,1/2]$ does the 1 come from the fact that it lies inside of $x^2+y^2=1$ ? Then I get $$\mathrm{d}S = \| \dot{r}_\rho \times \dot{r}_\phi \| = \sqrt{1 - \rho^2(\phi)} \rho \mathrm{d}\rho \mathrm{d}\phi$$ but this seems wrong. I end up with some elliptic integral, where is my mistake ? –  N3buchadnezzar May 17 '12 at 0:05
    
@N3buchadnezzar In the X-Y plane, $x^2+y^2=1$ is a circle with radius $1$, so $\rho$ changes from $0$ to $1$. –  Ayman Hourieh May 17 '12 at 0:07
    
@N3buchadnezzar As for $dS$, I think you made a mistake there. Can you edit the question to show how you derived that? –  Ayman Hourieh May 17 '12 at 0:13
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