Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=\frac{3}{x+1}, \quad y=0, \quad x=0, \quad x=8 $$

Short answer

The volume of the solid generated by revolving the region bounded by the graphs of \( y=\frac{3}{x+1}, y=0, x=0, x=8 \) about the \(x\)-axis is \(8\pi\) cubic units.

Step-by-step solution

Identify the revolved area

The region that is being revolved here is formed by the curve \(y=\frac{3}{x+1}\), \(x\)-axis (\(y=0\)), and the vertical lines \(x=0\) and \(x=8\). This is the area between the curve and \(x\)-axis, from \(x=0\) to \(x=8\).

Establish the disk volume formula

Our goal is to sum up the volumes of many tiny disks from \(x=0\) to \(x=8\). The volume of a disk with small thickness \(dx\) is given by \(dV = \pi [y(x)]^2 dx\), where \(y(x) = \frac{3}{x+1}\). The total volume \(V\) will be found by integrating \(dV\) from 0 to 8.

Setup the integration

Plug \(y(x) = \frac{3}{x+1}\) into the disk volume formula, we have: \(dV = \pi [\frac{3}{x+1}]^2 dx \). The total volume \(V\) is then given by the definite integral: \(V = \int_0^8 \pi [\frac{3}{x+1}]^2 dx \)

Compute the integral

The integration can be computed by using common techniques of integration. The integrand function can be simplified by squaring the fraction before the integration, resulting in \(V = \int_0^8 \pi \frac{9}{(x+1)^2} dx \). After integration, we get \( \pi [ -\frac{9}{x+1} ] |_0^8 = \pi (-1 - (-9)) = 8\pi \) cubic units.