Volume of revolution problem. Tricky

[nick.martinez]

Homework Statement

Find the volume of the solid rotated about the given axis

Homework Equations

∫R^2-r^2dx
Disk method

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.

 

[haruspex]

Pls give a full statement of the actual problem.

 

[nick.martinez]

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

 

[NascentOxygen]

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

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.

I'm using the disk method here.

No one can fault you on that.

 

[Ray Vickson]

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

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.

 

[NascentOxygen]

...

Please do not double post your questions! https://www.physicsforums.com/showthread.php?t=693391

 

[nick.martinez]

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.

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

 

[Ray Vickson]

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

I also get V = 16π/7.

 

[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.

 

[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 $$V = 2\pi \int_{-1}^0 -y dy + 2\pi \int_0^1 (1 + y) (1-y^{1/3})\,dy = \frac{23\pi}{14}$$