Triple Integral Troubles I'm having trouble calculating this integral. I can do the first one just fine, but it's in simplifying and calculating the third integral where I get stuck. 
$16\int_0^\frac{\pi}{4}\int_0^1\int_0^{\sqrt{1-r^2cos^2(\theta)}}rdzdrd\theta$
$16\int_0^\frac{\pi}{4}\int_0^1r\sqrt{1-r^2cos^2(\theta)}drd\theta$
with u-substitution:
$16\int_0^\frac{\pi}{4}\frac{-1}{3}(\frac{(1-cos^2(\theta))^(\frac{3}{2})}{cos^2(\theta)})d\theta$
I know the answer should be $16-8\sqrt{2}$

You're forgetting a "$-1$" term. Let $u = 1 - r^2\cos^2\theta$ so that $du = -2r\cos^2\theta \, dr$. Then observe that:
\begin{align*}
\int_0^1 r \sqrt{1 - r^2\cos^2\theta} \, dr
&= \frac{1}{-2\cos^2\theta}\int_1^{1-\cos^2\theta} \sqrt{u} \, du \\
&= \frac{1}{-2\cos^2\theta}\left[\frac{u^{3/2}}{3/2}\right]_1^{1-\cos^2\theta} \\
&= \frac{1}{-3\cos^2\theta}\left[(1-\cos^2\theta)^{3/2} - 1\right] \\
\end{align*}
Hence, we obtain:
$$
\frac{-16}{3}\int_0^{\pi/4} \frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta} \, d\theta
$$
Using Wolfram|Alpha, this integral does indeed equal $16 - 8\sqrt2$, as desired.

If you want to actually compute the antiderivative by hand, then here are some trig identities that will magically simplify things:
\begin{align*}
\frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta}
&= \frac{(\sin^2\theta)^{3/2} - 1}{\cos^2\theta} \\
&= \frac{\sin^3\theta - 1}{\cos^2\theta} \\
&= \frac{\sin\theta(1 - \cos^2\theta) - 1}{\cos^2\theta} \\
&= \frac{\sin\theta - \sin\theta\cos^2\theta - 1}{\cos^2\theta} \\
&= \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta}  - \sin\theta - \frac{1}{\cos^2\theta} \\
&= \sec\theta\tan\theta - \sin\theta - \sec^2\theta
\end{align*}
Thus, the antiderivative is:
$$
\sec\theta + \cos\theta - \tan\theta
$$
