# Integral Problem

1. Jun 8, 2006

### Metal

Hello, i'm new around here. I was having trouble with a problem, i thought i could look for help on the net. Anyway here's the problem:

Calculate the volume of a solid obtained by the rotation around Ox of all points (x,y) in RxR where y >= x*x, y <= square root of x and y <= 1/(8x).

What I did was:
1/(8x) = x*x so x = 1/2
1/(8x) = square root of x so x = 1/4

And then i found the integrals which are:
+ Integral of square root of x times dx between 0 and 1/4
- Integral of x * x times dx between 0 and 1/4
+ Integral of 1/(8x) times dx between 1/4 and 1/2
- Integral of x * x times dx between 1/4 and 1/2

So this gives me the area, but i have no idea on how to calculate the volume.
Hope someone can help.

Tks...
And i'm sorry about my english, this is my first time writing math in english.

2. Jun 8, 2006

### TD

You have found the correct points of intersection which divides your problem into to areas: from 0 to 1/4 and then from 1/4 to 1/2.

For a function f(x), the volume of the solid of revolution obtained by rotating f(x) about the x-axis between x=a and x=b is given by:

$$\pi \int\limits_a^b {f\left( x \right)^2 dx}$$

So you don't need the area -> no need to find the integrals you listed.
Do you think you can take it from here? If not: ask for help

3. Jun 8, 2006

### Metal

Ok, tks...
But why does this integral calculates this volume?

I mean... you took those integrals i made and multiplied each for its f(x) and for pi. Why does that make the volume?

Again sorry about any wrong english.

Last edited: Jun 8, 2006
4. Jun 8, 2006

### TD

If you revolve f(x) about the x-axis, you create 'circles' at each x-value with center (x,0) and radius f(x), perpendicular to the x-axis of course. To obtain the volume, you need to add all the areas of the discs, with the area given by pi*r² with r the radius. Here, r is f(x) so you integrate pi*f(x)² over the interval. Is that clear enough?

5. Jun 8, 2006

### Hootenanny

Staff Emeritus
Because, the formula rotates the 2D shape around the x axis. Imagine drawing a triangle form the orgin to some point x = b. Now imaging rotating that triangle around the x-axis. If you are having problems visualising this, draw it in a peice of paper, cut it out and rotate the triangle towards you. Now, if you follow the outline of this triangle, you have formed a 3D cone. The integral gives the volume enclosed by the cone.

Edit: Apologies for jumping in TD

Last edited: Jun 8, 2006
6. Jun 8, 2006

### TD

No problem at all!

7. Jun 8, 2006

### Metal

Oh I see... That was quite obvious actually... Tks.

8. Jun 8, 2006

### TD

You're welcome