Solving an Integral Problem: Find Volume of Solid

[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.

 

[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

 

[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.

 

[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?

 

[Hootenanny]

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 piece 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

 

[TD]

Edit: Apologies for jumping in TD

No problem at all!

 

[Metal]

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

 

[TD]

You're welcome