The first quadrant region enclosed by y=2x, the x-axis and the line x=1 is resolved about the line y=0. How do you find the resulting volume?

Question

Christopher Agresta · Accepted Answer

Evaluate #piint_0^1 (4x^2) dx# to get #V = (4pi)/3#

The region is shown in the graph below.

The graph shows a representative slice of the region. The slice is taken at some value of #x# and had width (or thickness) #dx#.

When rotated about the #x# axis, the slice generates a disk of radius #r = 2x#.

The volume of the representative disk is

#pir^2*"thickness"#.

In this case we have:

the volume of the slice is #pi(2x)^2 dx#.

#x# varies from #0# to #1#, so the solid has volume

#V = int_0^1 (pi(2x)^2) dx = pi int_0^1 4x^2 dx = (4pi)/3#

Note

Because the question was posted under the topic "Calculating Volumes using Integrals", I have used the integral.

With some imagination we might notice that the solid will be a right circular cone.
The cone will have radius #r = 2# and height #h = 1#. (It's on its side with the vertex or apex at the origin).

The volume of a right circular cone is

#V = 1/3 pir^2#.

Using this formula we also get

#V = (4pi)/3#.

Axel Alvarez · Answer

undefined To find the resulting volume, you can use the method of cylindrical shells. First, determine the limits of integration by finding the intersection point between y = 2x and x = 1, which is (1, 2). Then, the radius of each shell is the distance from the line of rotation (y = 0) to the axis of rotation (x = 1), which is 1. The height of each shell is given by the function y = 2x. Therefore, the integral for the volume is from x = 0 to x = 1 of 2πx(1)dx. Integrate this expression to find the volume.