Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(x\) -axis. $$ y=\sqrt{x+2}, \quad y=x, \quad y=0 $$

Short answer

The volume of the solid generated by revolving the region bounded by the curves \(y = \sqrt{x+2}, \quad y=x, \quad y=0\), about the \(x\)-axis is \(\frac{19\pi}{2}\)

Step-by-step solution

Identify the radius and height

For the shell method, the distance from the shell to the axis of rotation is our radius. Since we're rotating around the \(x\)-axis, our radius will be \(y\). The height of the shell will be the distance along the axis of rotation, which is \(x\). So here, \(y\) is the radial function and \(h = \sqrt{x+2} - x\) is the height function.

Identify the bounds of integration

To find the bounds in the shell method, we find where the functions intersect. So we solve \( y = x\) and \(y = \sqrt{x+2}\). After solving these two equations, we get \(x=1\) and \(x=4\). So our bounds of integration are \(1\) and \(4\).

Set up and solve the integral

Now we're ready to set up and solve the integral for the volume. So \(V = 2 \pi \int_{1}^{4} yh \, dx = 2\pi \int_{1}^{4}y(\sqrt{x+2} - x) \, dx\). After calculating this integral, we get \(V = 2\pi(\sqrt{2}[2\sqrt{2}+3]-\frac{9}{2}) = \frac{19\pi}{2}\).