# Solid of revolution

1. Aug 26, 2005

### ~angel~

Could someone please explain how to do this question.

Find the volume of the solid formed when the area between y=√x and y=x^4 is rotated about the line x=2.

I know how to do this when it's rotated about the x and y axes, but I'm not sure how to do it with a different line.

Thank you.

2. Aug 26, 2005

### AKG

Option 1: compute the area bounded by those curves, then find the x co-ordinate of its centroid and find the circumference of the circle whose radius is the distance from this centroid to the line x=2. Multiply the area by the circumference to get your volume (Pappus's Second Theorem).

Option 2: Notice why this problem is the same as the problem of finding the volume of the solid formed when the area between y = √(x + 2) and y = (x + 2)4 are rotated about the x-axis, and use this to solve the problem as you already know how to.

3. Aug 26, 2005

### Severian596

Do you mean about the y-axis? If the original problem says to rotate the solid about the line x=2.

4. Aug 26, 2005

### AKG

Yes, you're right, the y-axis (the line x = 0).

5. Aug 28, 2005

### ~angel~

6. Aug 29, 2005

### ~angel~

I can't seem to get the answer. I'm using the 2nd method. I'm meant to be rotating about the y-axis, aren't I?

In that case,

x^2=(y^2-2)^2------------- x^2=((y^1/4)-2)^2
=y^4-4y^2+4 --------------= y-(4y1/4)+4

Then you integrate these? The bounds are y=0, y=1. But I end up with completely the wrong answer.

Last edited: Aug 29, 2005
7. Aug 29, 2005

### TD

By using $y = \sqrt {x+2}$ and $y = (x+2)^4$, you 'moved' the functions two places to the left. Your bounds have to move then as well, so you integrate from y = -2 to y = -1.

The volume of a solid of revolution, about the y-axis, is the given by:

$$\pi \int\limits_a^b {f\left( y \right)^2 dy}$$

8. Aug 29, 2005

### ~angel~

Okay, the bounds move, but that results in an undefined answer, if my expansion above is correct.

Shouldn't it be y= 2,1. You can't get a negative answer from those equations.

9. Aug 29, 2005

### TD

Your expansion isn't really clear to me...

10. Aug 29, 2005

### ~angel~

I'm not sure how I can make it clearer, but when you rearrange $y = \sqrt {x+2}$, you get $x^2 = y^4-4y^2+4$.

When you rearrange $y = (x+2)^4$, you get $x^2 = y-4y^(1/4)+4$

For the latter, that is the 4th root for the 2nd term.

11. Aug 29, 2005

### TD

If you have 2 functions g(x) and h(x) and you want to find the solid of revolution of their difference (f(x) = h(x) - g(x)), the function f(x)² in the previous mentioned formula isn't h(x)²-g(x)² but (h(x) - g(x))² which isn't the same.

12. Aug 29, 2005

### ~angel~

okay, $y = \sqrt {x+2}$, is rearranged to $x = y^2-2$.

$y = (x+2)^4$ is rearranged to $x = y^(1/4)-2$

Then, $[(y^2-2)-(y^(1/4)-2)]^2$

Then you get $y^4-y^2+y^(1/16)$

But shouldn't you just integrate from the same bounds because your only changing the values of the x-axis.

13. Aug 29, 2005

### TD

I'm sorry, I think I confused you and my last answer wasn't entirely correct.

Using f(x) = h(x) - g(x) doesn't change the area, but it does move it so the volume of the solid of revolution doesn't stay the same. Therefore, it would be indeed better to calculate the entire V of the solid of revolution of h(x) and then substract the one of g(x).