# Volume of revolution problem. Tricky

1. May 24, 2013

### nick.martinez

1. The problem statement, all variables and given/known data
Find the volume of the solid rotated about the given axis

2. Relevant equations
∫R^2-r^2dx
Disk method

3. The attempt at a solution
I'm having trouble finding the limits of integration: here's my setup
Pi∫(x^3+1)dx and integrated from -1 to 1. I got this by setting y=-1 equal to y=x^3
X^3=-1=-1

Then I get (16/7)pi which I believe is the right answer so can anyone tell me whether my method is correct. I'm using the disk method here.

2. May 24, 2013

### haruspex

Pls give a full statement of the actual problem.

3. May 24, 2013

### nick.martinez

Y=x^3 ; y=-1 x=1 ; axis; y=-1

4. May 24, 2013

### Staff: Mentor

As you show no working, don't include a diagram, and haven't adequately related the question, it is not possible to know what your method is. Nevertheless, after some faltering attempts at telepathy, a bit of guesswork and reading between the lines, I too arrived at an answer of 16Pi/7.

But of course, that is for my question, and as yet we have no way of knowing whether my question is similar to yours.

No one can fault you on that.

5. May 24, 2013

### Ray Vickson

This is meaningless. What axis? Do you mean that you want the region bounded by y = -1, y = x^3 , x = 1 and some axis? Please state the problem properly.

6. May 24, 2013

### Staff: Mentor

7. May 24, 2013

### nick.martinez

Yes ray vickson those are the bounds and it is being rotated about the y=-1. It would be appreciated.

8. May 25, 2013

### Ray Vickson

I also get V = 16π/7.

9. May 25, 2013

### CAF123

nick.martinez,
From looking at this post, it appears because of the initial ambiguity in the problem statement that I misinterpreted what the region was when I replied in the other thread. I too get 16pi/7 and there is no inner radius here because the region required extends all the way to the line of rotation.

10. May 25, 2013

### CAF123

On second thoughts, I don't think I do get $16 \pi/7$. See the following: via the disk method, the radius of a disk is $(1 + x^3)$ hence this gives a volume of $$V = \pi \int_0^1 (1 + x^3)^2\,dx = \frac{23\pi}{14}$$ $x$ varying from 0 to 1.

Similarly, via the cylindrical shell method: split whole region into two separate regions, since shells will be of varying height. Then we have $$V = 2\pi \int_{-1}^0 -y dy + 2\pi \int_0^1 (1 + y) (1-y^{1/3})\,dy = \frac{23\pi}{14}$$