Question

What is the volume of the solid generated when the region in the first quadrant bounded by the graph of y=x^3, the x-axis, and the vertical line x=2 is revolved about the x-axis?
Show work.

A
$4$
B
$\frac{128}{7}$
C
$4\pi$
D
$\frac{128\pi}{7}$
​

Answer 1

Answer:

The first thing we need to do is to find the area bounded by:

y = x^3

y = 0

between:

x = 0 and x = 2

This is the integral of the given function between x = 0 and x = 2, written as:

$\int\limits^2_0 {x^3} \, dx = \frac{2^4}{4} - \frac{0^4}{4} = 2^2 = 4$

This means that the area of the bounded region is 4 square units.

Now, if we do a full rotation around the x-axis, the volume generated will be equal to the area that we obtained times 2*pi units.

The volume is:

V = (4 square units)*(2*pi units) = 8*pi cubic units.

(Notice that no option coincides with this, there may be a mistake in the options)