Volume of a solid revolution

Question

Find the volume of a solid figure generated by rotating the area of the region bounded by, $$y_1=x^2-4$$ $$y_2=3x+6$$ and the $x$-axis about the $x$ axis.

I tried solving this using the formula, $$V= \pi \int_{-2}^{5}y_2^2-y_1^2 dx$$

Answer 1

Notice that for x from -2 to -1, $y_1$ is dominating. For x from -1 to 2, $y_2$ is dominating. For x from 2 to 5, $y_2 - y_1$ is dominating.

Hence the volume is

$$\int_{-2}^{-1}\pi (x^2 - 4)^2dx + \int_{-1}^{2}\pi (3x + 6)^2dx + \int_{2}^{5}\pi ((3x + 6)^2 - (x^2 - 4)^2)dx$$

$$= 366\pi$$