# Volume of the resulting solid

1. Oct 25, 2004

### Caldus

I am having a lot of trouble with these last 3 problems out of 10 that I have done relating to volumes. I have tried just about every method for each of these and I am just not getting the right answer (according to this online math program I am using).

2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:
y = x^2, y = 1; about y = 2.

The resulting solid for me looks like a volcano. So I used the Shell method like so:

A(x) = length * width
A(x) = 2*pi*(x)*x^2 = 2*pi*x^3

V = integral b/w 0 and 1 of A(x) ...

Not getting the right answer here either...

3. A ball of radius 11 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

I tried using vertical washers like so:

A(x) = pi*11^2 - pi*4^2

V = integral b/w 1 and 11 of A(x) = ...

Incorrect here as well. Not sure how else to approach this one.

Thanks for help on any of these. At least I got the other 7 problems done on my own. :P

Last edited: Oct 26, 2004
2. Oct 26, 2004

### Galileo

$40x-8x^2$ is a parabola with it's maximum at x=2.5. So the region bounded by this parabola and the line y=0 is like a hill.
When you rotate it around the y-axis you get something like a donut, but with the upper part like a parabola and the bottom part removed, so it's flat on the bottom side.
Using horizontal disks is right, but you have to integrate from y=0 to y=50. (From bottom to top).
Each horizontal disk is like a ring. Try setting up the integral from here.

3. Oct 26, 2004

### Caldus

Thanks, I eventually figured it out.

Still stuck on the other two problems...

4. Oct 26, 2004

### arildno

In the second, I would suggest that you split the volume up in the following manner:

The volum of this cylinder is subtracted from:

Outer Volume: Radius:$$2-x^{2}$$

$$V=\int_{0}^{1}\pi((2-x^{2})^{2}-1^{2})dx$$

5. Oct 26, 2004

### Caldus

I ended up getting 5.864306287 as the answer but the online homework program says it's still incorrect.

I agree that the small radius is 1. How did you come up with 2 - x^2 as the big radius though? I don't get how to determine what the outer radius is in this case. Thanks for any help.

Any ideas on the last problem?

6. Oct 26, 2004

### Caldus

OK, got an idea for the last problem but still not getting the right answer.

OK so if I place a sphere of radius 11 in the center of the x-y plane, then the top right quadrant curve of that sphere will equal (r^2 - x^2)^(1/2) correct? So then I can do the volume like so (using vertical washers):

V = 2pi * Integral b/w 0 and 11 of (11^2 - x^2) - (4^2 - x^2) dx

But it isn't right according to this program.

7. Oct 26, 2004

### arildno

You are to rotate it around y=2, right?
The distance from y=2 to the parabola is 2-x^2

8. Oct 26, 2004

### Caldus

Yeah I suppose so. Grrr this stupid program. It must be a bug or something...

9. Oct 26, 2004

### Galileo

For the last problem, you want to rotate the area bounded by $y=\sqrt{11^2-x^2}$ and $y=2$ about the x-axis.

The answer I get is $\frac{4268\pi}{3}\approx 4469.44$

10. Oct 26, 2004

### Caldus

Wouldn't it be y = 4 and not y = 2 for the other one?

I tried your answer and it did not work while I tried my answer using y = 4 and of course it says incorrect.

I'm running out of ideas as to what I could have done wrong for both of these problems.

11. Oct 26, 2004

### Caldus

Anyone? These problems are really frustating me...

12. Oct 27, 2004

### Galileo

Yes it is. Sorry for that. It should be 4506.84 (if I did it right this time).
Does that confirm?

Also, the integral Arildno gave should run from -1 to 1, so the answer will be doubled.

13. Oct 27, 2004

### arildno

AARGH!
That's where my mistake was!
Thx, Galileo, sorry Caldus..