Volume bounded by two hyperboloids

If we look at the boundaries for the outer function for positive z (due to symmetry, we can multiply by 2 to get the total volume), we get

$\begin{array}{lcl} 0\leq r\leq\sqrt{z^{2}+1}\\ 0\leq \theta \leq 2\pi \\ 0\leq z \leq5 \end{array}$

And for the inner function, we get

$\begin{array}{lcl} 0\leq r\leq\sqrt{z^{2}-1}\\ 0\leq \theta \leq 2\pi \\ 1\leq z \leq5 \end{array}$

Then we get this expression for the total volume:

$2\left[\int_{0}^{5}\int_{0}^{2\pi}\int_{0}^{\sqrt{z^{2}+1}}r \,drd\theta dz-\int_{1}^{5}\int_{0}^{2\pi}\int_{0}^{\sqrt{z^{2}-1}}r \,drd\theta dz \right]$

$=\int_{0}^{5}\int_{0}^{2\pi} z^{2}+1\,d\theta dz-\int_{1}^{5}\int_{0}^{2\pi} z^{2}-1\,d\theta dz$ $=\int_{0}^{5}\int_{0}^{2\pi} z^{2}+1\,d\theta dz-\int_{1}^{5}\int_{0}^{2\pi} z^{2}-1\,d\theta dz$

$=\int_{0}^{5} (2\pi)(z^{2}+1)\,dz-\int_{1}^{5} (2\pi)(z^{2}-1)\,dz$

$= (2\pi)\left(\frac{z^{3}+1}{3}+z\right) \Big|^5_0 - (2\pi)\left(\frac{z^{3}+1}{3}-z\right)\Big|^5_1$

$=\frac{56\pi}{3}$

$a+b=56+3=59$

Comparing the volumes of revolution, we want to calculate $\pi\int_{-5}^5 (z^2 + 1)\,dz - 2\pi\int_1^5 (z^2 - 1)\,dz \; = \; \tfrac{56}{3}\pi$ making the answer $56+3=\boxed{59}$ .