Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(y\) -axis. $$ y=9-x^{2}, \quad y=0, \quad x=2, \quad x=3 $$

Short answer

The volume of the solid generated by revolving the region bounded by the given equations about the y-axis is \(21\pi\) units

Step-by-step solution

Identify the limits of integration

From the given bounds, observe that the region is from \(x = 2\) to \(x = 3\). Therefore, these are the limits of integration.

Set up the integral

The volume \(V\) of the solid of revolution generated by rotating the region bounded by \(y=9-x^{2}\) and \(y=0\) about the \(y\)-axis can be found using the formula for the disk method: \(V = \pi \int_{a}^{b} r^{2} dx\), where \(r\) is the radius of the disc (the distance from the disc to the axis of rotation, in this case \(y\)), and \(a\) and \(b\) are the limits of integration. Here, \(r = \sqrt{9-x^{2}}\). Thus, \(V = \pi \int_{2}^{3} (\sqrt{9 - x^{2}})^{2} dx\). Simplify the integrand to obtain \(V = \pi \int_{2}^{3} (9 - x^{2}) dx\).

Evaluate the integral

Perform the integral: \(V = \pi [9x - \frac{x^{3}}{3}]\_{2}^{3}\).

Calculate the volume

Substitute the limits into the expression to find the volume: \(V = \pi (9*3 - \frac{3^{3}}{3}) - \pi (9*2 - \frac{2^{3}}{3})\). Simplify this to get \(V =\pi(21)\).