The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. 
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
  $$y^2 − x^2 = 9, \quad y = 4; \text{ about the $x$-axis}$$

I tried using the washer method in terms of $x$ and I got the wrong answer. Here is how I set up my integral
$$V=\int_3^4 (4)^2-(\sqrt{y^2+9})^2 = 76\pi/3$$
 A: I think your limits of integration are incorrect. If you substitute $y = 4$ into $y^2 - x^2 = 9$, you find that $x = \pm \sqrt{7}$. Therefore, the two curves intersect at $x = \pm \sqrt{7}$. By washer method, we have:
$$\begin{align}
V &= \pi \int_{-\sqrt{7}}^{\sqrt{7}} (4)^2-(\sqrt{x^2+9})^2 \,dx \\
&= 2\pi \int_{0}^{\sqrt{7}}16 - (x^2+9) \,dx \\
&= 2\pi \int_{0}^{\sqrt{7}}7 - x^2 \,dx \\
&= 2\pi \left[ 7x - \frac{1}{3}x^3 \right]_{0}^{\sqrt{7}} \\
&= 2\pi \left( \frac{14\sqrt{7}}{3} \right ) \\
&= \frac{28\pi\sqrt{7}}{3}
\end{align}$$
And just for fun, let's try the shell method. Here, we have no choice but to find the volume obtained by revolving just the part of the region in the first quadrant, and doubling it.
$$\begin{align}
V &= 2 \times 2\pi \int_{3}^{4}  y\sqrt{y^2 - 9}\,dy \\
V &= 4\pi \int_{3}^{4}  y\sqrt{y^2 - 9}\,dy \\
V &= 4\pi \left[ \frac{1}{3}(y^2 - 9)^{\frac{3}{2}}\right]_{3}^{4} \\
V &= 4\pi \left[ \frac{1}{3}(y^2 - 9)^{\frac{3}{2}}\right]_{3}^{4} \\
V &= 4\pi \left[ \frac{7\sqrt{7}}{3} \right ] \\
V &= \frac{28\pi\sqrt{7}}{3}
\end{align}$$
