# Just wondering if I set this up oK

## Homework Statement

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

## Homework Equations

y = x^2 +1 , y = 9-x^2: about y = -1

## The Attempt at a Solution

I used disks. I said that r = 1 + ((9-x^2)-(x^2 +1 )) = 9 - 2x^2
So my integral all together is

V = ∏ ∫ (9 - 2x^2)^2 dx from [-2,2]

Does this set up look correct.???

Also I have another one I would like to confirm.
This one your supposed to set up the integral not evaluate for the volume of a solid.
y = tanx, y = x, x = pi/3 : about the y axis.
I have
V = 2∏∫ x(tanx - x) dx between 0 and ∏/3
I think it is good but just to be safe.

Mark44
Mentor

## Homework Statement

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

## Homework Equations

y = x^2 +1 , y = 9-x^2: about y = -1

## The Attempt at a Solution

I used disks.
Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.

Disks won't work, but washers will. A washer is a disk with a hole in its middle. For a typical volume element, the outer radius goes from the line y = -1 to the curve y = 9 - x2. The inner radius goes from the line y = -1 to the curve y = x2 + 1.
I said that r = 1 + ((9-x^2)-(x^2 +1 )) = 9 - 2x^2
So my integral all together is

V = ∏ ∫ (9 - 2x^2)^2 dx from [-2,2]

Does this set up look correct.???

Mark44
Mentor
Also I have another one I would like to confirm.
This one your supposed to set up the integral not evaluate for the volume of a solid.
y = tanx, y = x, x = pi/3 : about the y axis.
I have
V = 2∏∫ x(tanx - x) dx between 0 and ∏/3
I think it is good but just to be safe.
Looks OK to me. It would have been helpful to indicate that your using shells.

Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.

Disks won't work, but washers will. A washer is a disk with a hole in its middle. For a typical volume element, the outer radius goes from the line y = -1 to the curve y = 9 - x2. The inner radius goes from the line y = -1 to the curve y = x2 + 1.

Yeah I always draw it. Sorry for the confusion with the other problem not indicating I was using shells.

r = 1 + ( x^2 + 1) R = 1+ (9-x^2)
So V = ∏∫( x^2 +2)^2 - (10-x^2)^2 dx I think that is it but I have a little trouble with the radius sometimes.

Last edited:
Mark44
Mentor
Yeah I always draw it. Sorry for the confusion with the other problem not indicating I was using shells.

r = 1 + ( x^2 + 1) R = 1+ (9-x^2)
So V = ∏∫( x^2 +2)^2 + (10-x^2)^2 dx I think that is it but I have a little trouble with the radius sometimes.
Your radii are correct but the integral isn't. The volume of a washer is ##\pi(R^2 - r^2)(\text{thickness})##.

I just mean minus. Check it out I edited.

V = ∏∫(10-x^2)^2 -( x^2 +2)^2 dx GRR.

Mark44
Mentor
That's better.

It is like I'm thinking one thing and typing another. Thanks for the confirmation.