Area of a portion of a paraboloid

**arbolis** · August 4th 2009, 06:22 PM

Find the area of the portion of the paraboloid $x=y^2+z^2$ which is inside the cylinder $y^2+z^2=9$ .

My attempt : I notice that to be more friendly, I could change the problem as "the paraboloid $x^2+y^2=z$ and the cylinder $x^2+y^2=9$ ", but I won't do that.
So $A=\iint _S \sqrt{ \left ( \frac{ \partial f}{\partial y} (y,z) \right ) ^2+ \left ( \frac{\partial f}{\partial z} (y,z) \right ) ^2+1}$ $=\iint _S \sqrt {4(y^2+z^2)+1}dydz$ .
The 2 surfaces cross each other when $x=9$ . So when $y^2+z^2=9$ .
I converted the variables into polar coordinates, $A=\int _0^{2\pi} \int _0^{9} \rho \sqrt{4\rho +1} d\rho d\theta$ .
Is that good?

**Haytham** · August 5th 2009, 01:38 AM

Shouldn't it be

$\rho \sqrt{4\rho^2 +1}$ ?

and the $\rho$ integration limit between 0 and $\sqrt{9}$

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