Question

In Exercises 29-32, find the volume of the solid described. Find the volume of the solid generated by revolving the triangular region bounded by the lines \(y=2 x, y=0,\) and \(x=1\) about (a) the line \(x=1\) (b) the line \(x=2\)

Short answer

The volume of the solid generated by revolving the triangular region bounded by the lines \(y=2x\), \(y=0\), \(x=1\) about (a) the line \(x=1\) is \(\frac{1}{3}\pi\) cubic units and (b) the line \(x=2\) is \(\frac{4}{3}\pi\) cubic units.

Step-by-step solution

Identify the region

First, it necessary to understand the region that is being revolved. Since one of the boundaries is \(y=0\), this means the triangle's base lies on the x-axis and goes from \(x=0\) to \(x=1\). The line \(y=2x\) forms the hypotenuse of the triangle.

Setup the integral for x=1

When revolving the region around the line \(x=1\), the solid generates a cone shape with height and radius both equal to 1. The volume of such a solid can be calculated with the formula for the volume of a cone which is \(\frac{1}{3} \pi r^2 h\). Substituting \(r=1\) and \(h=1\), volume becomes \(\frac{1}{3}\pi\).

Setup the integral for x=2

When revolving around the line \(x=2\), the solid generates a shape known as a washer. To find the volume, we can apply the method of Disk/Washer method. The radius of the large disk is \(r_1 = 2-0 = 2\), and small disk is \(r_2 = 2 - 2x\), where \(x\) is between \(0\) and \(1\).By applying the Washer method, the volume is given by \(\int_a^b \pi [f(x)]^2 dx - \int_a^b \pi [g(x)]^2 dx = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) dx\). Substituting for \(a=0, b=1, f(x)=2\), and \(g(x) = 2- 2x\), the volume becomes \(\pi \int_0^1 ([2]^2 - (2 - 2x)^2) dx\).

Evaluate the Volume

Evaluating the integral, the volume for the solid obtained by rotating around \(x=2\) is \(V=\frac{4}{3}\pi\).