The region bounded by the graphs of y =√x and y = x^2 is rotated about the line x = 2. Find the volume of the solid that is generated.

Question

Not sure where to start. Thank you in advance!

Answer 1

This Desmos graph has pictures for both the shell and washer methods. Enable Shell and Washer images in turn.

desmos.com/calculator/gsdylvkdwc

This Desmos graph is the graph from where the images in the previous graph were generated. Enable and expand Shell and Washer folders to see the explanation for each. You can use the movable points to see different positions for outer/inner radius (washer) or shell walls:

desmos.com/calculator/2ryatmnjai

Here is a Desmos 3d graph that shows the solid being generated. Use the slider on k from 0 to 2pi to see the creation.

desmos.com/3d/bfo6im38ik

The following is an explanation for using the shell method.

The points of intersection for the two functions are at (0,0) and (1,1).

The shell walls will appear from x = 0 to x = 1. The height of a typical shell wall is y-coordinate at the top minus y-coordinate at the bottom or: √x - x². The distance from the axis of revolution (x = 2) to the center of a typical shell wall is x-coordinate at the right minus x-coordinate at the left or: 2 - x.

Creating the definite integral leads to:

2π ∫₀¹ (x^(1/2) - x²)(2 - x)dx

Let's simplify the integrand (using FOIL):

2x^(1/2) - x^(3/2) - 2x² + x³

The antiderivative (without showing the constant of integration):

2(2/3)x^(3/2) - (2/5)x^(5/2) - (2/3)x³ +(1/4)x⁴

Since we are evaluating from 0 to 1:

[(4/3 - 2/5 - 2/3 + 1/4) - (0)]

80/60 - 24/60 - 40/60 + 15/60

31/60

But do not forget to multiply by 2π:

2 π (31/60) = 31π/30