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

My attempt : I notice that to be more friendly, I could change the problem as "the paraboloid $\displaystyle x^2+y^2=z$ and the cylinder $\displaystyle x^2+y^2=9$", but I won't do that.

So $\displaystyle 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}$$\displaystyle =\iint _S \sqrt {4(y^2+z^2)+1}dydz$.

The 2 surfaces cross each other when $\displaystyle x=9$. So when $\displaystyle y^2+z^2=9$.

I converted the variables into polar coordinates, $\displaystyle A=\int _0^{2\pi} \int _0^{9} \rho \sqrt{4\rho +1} d\rho d\theta$.

Is that good?