# Volumes of Revolution Question

• RJLiberator
Great job summarizing the conversation! In summary, the conversation discusses computing the volume of a solid of revolution obtained by rotating a region bounded by two curves around the axis y=-1. The disc method is used and the solution involves calculating the volume of the entire region from 0 to 1 using the radius of x^2+x+1, then subtracting the volume of the squared middle region with a radius of 1. The final answer is 27pi/10.

Gold Member

## Homework Statement

Let R be the region bounded by the curves y=0 and y=x^2+x between x=0 and x=1. Compute the volume of the solid of revolution obtained when R is rotated about the axis y=-1.

## Homework Equations

Disc method: integral of pi*r^2 = volume from the bounds

## The Attempt at a Solution

The bounds are obvious = 0 to 1.

The hard part is accounting for the empty area inbetween as we rotate around y=-1 rather then y=0.

What I did to solve the problem was moved the volume up 1 to the origin. I then solved the entire volume of the problem from 0 to 1 using the radius of x^2+x+1 (the +1 to account the entire area).
My answer here was 37pi/10.

I then went back and took the volume of the squared middle region which had radius of 1 and then subtracted that from my first answer.

My final answer was 27pi/10.

Can anyone confirm my train of thought/answer?

Thank you.

RJLiberator said:

## Homework Statement

Let R be the region bounded by the curves y=0 and y=x^2+x between x=0 and x=1. Compute the volume of the solid of revolution obtained when R is rotated about the axis y=-1.

## Homework Equations

Disc method: integral of pi*r^2 = volume from the bounds

## The Attempt at a Solution

The bounds are obvious = 0 to 1.

The hard part is accounting for the empty area inbetween as we rotate around y=-1 rather then y=0.

What I did to solve the problem was moved the volume up 1 to the origin. I then solved the entire volume of the problem from 0 to 1 using the radius of x^2+x+1 (the +1 to account the entire area).
My answer here was 37pi/10.

I then went back and took the volume of the squared middle region which had radius of 1 and then subtracted that from my first answer.

My final answer was 27pi/10.

Can anyone confirm my train of thought/answer?

Thank you.
That works for me.

You could also do the problem without translating the region upward like so:
##\Delta V = \pi[(\text{outer radius})^2 - (\text{inner radius})^2]\Delta x##
where the outer radius is ##x^2 + x - (-1) = x^2 + x + 1## and the inner radius is 1.
The volume integral is then
##\pi \int_0^1 [(x^2 + x + 1)^2 - 1^2] dx##

RJLiberator
Ah, so you managed to get the same answer as well? Confirming my result :D