TITLE: Volume of a solid revolution
QUESTION [0 upvotes]: 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,
$$V= \pi \int_{-2}^{5}y_2^2-y_1^2  dx$$

REPLY [0 votes]: 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
$$\int_{-2}^{-1}\pi (x^2 - 4)^2dx + \int_{-1}^{2}\pi (3x + 6)^2dx + \int_{2}^{5}\pi ((3x + 6)^2 - (x^2 - 4)^2)dx$$
$$= 366\pi$$
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