Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line \(x=6\). $$ x y=6, \quad y=2, \quad y=6, \quad x=6 $$

Short answer

The volume of the solid generated is \(9\pi\) units.

Step-by-step solution

Sketch the region

Start by sketching the region bounded by the given equations. This region is bounded by the rectangular hyperbola \(xy=6\), the lines \(y=2\), \(y=6\), and \(x=6\). Sketch all of these curves and identify the region.

Identify the radii for the washer method

The solid generated by revolving the region around the line \(x=6\) has the shape of a donut or washer, so use the washer method. The outer radius \(R(x)\) of the washer is the distance from the line \(x=6\) to the line \(y=6\), which is constant and equal to 6. The inner radius \(r(x)\) is the distance from the line \(x=6\) to the curve \(xy=6\), so \(r(x)=6-x\). This is because for points on the curve \(xy=6\), the y-coordinate is equal to \(6/x\).

Set up the integral

The volume \(V\) of the solid is given by the integral \(V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] dx\). In this case, \(a=3\), \(b=6\), \(R(x)=6\), and \(r(x)=6-x\). Therefore, the integral becomes \(V = \pi \int_{3}^{6} [(6)^2 - (6-x)^2] dx\).

Evaluate the integral

Evaluate the integral by first simplifying the integrand to get \(V = \pi \int_{3}^{6} [36 - (36 - 12x + x^2)] dx = \pi \int_{3}^{6} (12x - x^2) dx\). Then calculate the antiderivative at the endpoints to find \(V = \pi [6x^2 - x^3/3]_{3}^{6} = \pi [(6(6)^2 - (6)^3/3) - (6(3)^2 - (3)^3/3)] = \pi [72 - 72 -18 + 9] = \pi [18 - 9] = 9\pi\).