Question

Let G be the region enclosed by the curves y=4/x and y=5-x
Find the volume of the solid generated by rotating the region G about the x-axis.

Answer 1

1. Let $G$ be the region enclosed by the curves $y = 4 / x$ and $y = 5 - x$

Find the volume of the solid generated by rotating the region $G$ about the $x$ -axis.

Answer :

Here's the picture of the region G:

The intersections of the two curves are at (1,4) and (4, 1).

For revolution about the x-axis, we use the method of washers. The outside radius is the top curve $\frac{x}{4}$ and the inside radius is the bottom curve $5 - x$ . Thus, the volume V is given by

$$V = \pi \int_{1}^{4} \left[ (5 - x)^{2} - \left(\frac{4}{x}\right)^{2} \right] dx = \pi \int_{1}^{4} \left[ (5 - x)^{2} - \frac{16}{x^{2}} \right] dx$$ $$= \pi \left(- \int_{1}^{4} (5 - x)^{2} \, d(5 - x) - 16 \int_{1}^{4} \frac{1}{x^{2}} \, dx\right) = \pi \left\{ - \left[ \frac{(5 - x)^{3}}{3} \right]_{1}^{4} - 16 \left[ \frac{1}{x} \right]_{1}^{4} \right\} = 9\pi$$

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