# Calc II: Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.

I've been working on with the area of the region in my calc II class, and now have to deal with the volume.

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. $$y = 7\sqrt{49 − x^2}, y = 0, x = 2, x = 3$$ about the x-axis

I was able to draw the graph, but can't set up the integral. Any help would be appreciated. Thanks!

$$\pi \int_2^3 \left(7\sqrt{49 - x^2}\right)^2 \ dx$$
If you recall the disk method, the way we visualize finding the volume of the solid of revolution around the x-axis, is, we are essentially adding up infinitely thin disks of volume $dV$ for each small width $dx$ at each $x$ (a way to think of the integral). A disk at position $x$ will have radius equal to the distance from the x-axis to the curve, in this case $7\sqrt{49-x^2}$, and will have width $dx$, an infinitely small piece of length on the x-axis. Therefore $$dV = \pi r^2 h = \pi (7\sqrt{49-x^2})^2 \, dx = 49\pi(49-x^2) \, dx$$ and so \begin{align} V &= 49\pi \int_2^3 (49-x^2) \, dx \\ &= 49\pi \left[ 49-x^2 \right]_2^3 \\ &= 49\pi \left[ 49x-\frac{x^3}{3} \right]_2^3 \\ &= 49\pi \left[ 49-\frac{19}{3} \right] \\ &= 49\pi \cdot \frac{128}{3} \\ &= \frac{6272\pi}{3} \end{align}