Question

In Exercises \(47-50\), the integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$ 2 \pi \int_{0}^{2} x^{3} d x $$

Short answer

The plane region is the area enclosed by \(f(x) = x^{3}\), the x-axis, and the lines \(x=0\) and \(x=2\). The axis of revolution is the x-axis.

Step-by-step solution

Identify the function and limits

We are given the integral \(2 \pi \int_{0}^{2} x^{3} d x\). Here, the function \(f(x)\) is \(x^{3}\) and the limits of integration are \(a = 0\) and \(b = 2\).

Identify the plane region

The plane region is the region bound by the curve of the function \(f(x) = x^{3}\), the x-axis (the line \(y=0\)), and the lines \(x=a=0\) and \(x=b=2\). This is a region in the first quadrant, under the curve \(f(x) = x^{3}\), and between \(x=0\) and \(x=2\).

Identify the axis of revolution

From the form of the integral, we can infer that the region is revolved around the x-axis to form the solid of revolution. This is because the integral is in the form of the Disk Method, which is typically used to find volume when the solid is revolved around the x-axis.