$\displaystyle |\nabla f * p|=2z$

So, now I have

$\displaystyle \displaystyle \iint_{R} \frac{2\sqrt{2}}{2z}dA$

After factoring out constants and rewriting z as a function of x and y I have

$\displaystyle \displaystyle \sqrt{2}\iint_{R}\frac{1}{\sqrt{25-x^2-y^2}}dA$

Now, converting to polar coordinates, I get

$\displaystyle \displaystyle \sqrt{2}\int_{0}^{2\pi}\int^{4}_{0}\frac{1}{\sqrt{ 25-r^2}}rdrd\theta$

Ok, I let $\displaystyle u=25-r^2$. Then, $\displaystyle du=-2rdr$

Which then gives me

$\displaystyle \displaystyle \sqrt{2}\int_{0}^{2\pi}\frac{1}{2}\int^{4}_{0}\fra c{-1}{\sqrt{u}}dud\theta$

Then I get

$\displaystyle \displaystyle \sqrt{2}\int_{0}^{2\pi}[ -\sqrt{25-r^2}}]^{4}_{0} d\theta$

Now, I have

$\displaystyle \displaystyle \sqrt{2}\int_{0}^{2\pi}2d\theta$

So

$\displaystyle \frac{\sqrt{2}}{2}[\theta]^{2\pi}_{0}$

$\displaystyle =\frac{\sqrt{2}}{2}(2\pi)$

The answer in the book is $\displaystyle 20\pi$. can someone help me find my mistake(s)? Thanks.