Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

Question

1. y=6-x², y=2; about the x-axis
2. x=2-y², x=y⁴; about the y-axis
3. y=x³, y=1, x=2; about y=-3
4. y=x², x=y²; about x=-1

I can sketch the graph but I'm not sure how to proceed with the calculation of the volume.

Answer 1

In your first problem, let's find the region bounded by the two curves

y = 6 - x²

y = 2

I hope this region does not extend infinitely far to the left and right! Fortunately, the region ends at the two points where the curves meet:

6 - x² = 2

Solving 6 - x² - 2 = 0,

or 4 - x² = 0,

the left and right points (the "limits of integration") are at x = -2 and x = 2.

Let A be the volume of the solid obtained by rotating the region from x = -2 to x = 2 bounded by the two curves y = 6 - x² and y = 0 around the x axis. This volume is

A = π ∫(6 - x²)² dx, integrated from x=-2 to x=2.

To do this integration,

π∫(6 - x²)² dx = ∫(36 - 12x² - x⁴)dx = 36x - 4x³ - (1/5)x⁵ + a constant of integration

Integrating from x = -2 to x = 2, A = π ∫(6 - x²)² dx = π464/5.

Let B be the volume of the solid obtained by rotating the region from x = -2 to x = 2 bounded by the two (very straight) curves y = 2 and y = 0 around the x axis. This volume (a simple cylinder) is

B= π ∫2² dx, integrated from x=-2 to x=2.

To do this integration,

∫2² dx = ∫4 dx = 4x + a constant of integration

Integrating from x = -2 to x = 2, B = π ∫2² dx = π16.

(Glad that this agrees with the formula for the volume of a cylinder, V = base times height = 4π times 4.)

Volume A is a bagel without a hole. Volume B is a cylinder.

Subtract volume B from volume A and you now have a bagel with a hole.

The bagel with a hole is the volume that the problem asks for.

The volume of this bagel with a hole is A - B = (464/5)π - 16π = (384/5)π