Find the volume of the solid formed by rotating the region

1. Apr 28, 2008

zcabral

1. The problem statement, all variables and given/known data
Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^2
y=2x

2. Relevant equations

V= pi* integral [a,b] r^2

3. The attempt at a solution

So i used intergral 0 to 2 (2x-x^2)^2 dx
this is wat i got....
pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))
its wrong so wat did i do wrong?

2. Apr 28, 2008

dynamicsolo

You are using the "disk method", yes? But keep in mind that your disk has a hole in it: the area is not (pi)(2x - [x^2])^2 , but rather the area of the outer circle minus the area of the inner circle.

3. Apr 28, 2008

zcabral

but it doesn't technically have a hole that goes all the way through rite? wouldn't that still make it a disk? i know it has a like a big chunk out of it

4. Apr 28, 2008

dynamicsolo

Every slice of the volume will have a hole in it, so each slice is a ring (or annulus). The area is going to be the area of the disk defined by the outer radius minus the hole which has the inner radius. That difference is not given by the expression you are using. How should it be written?

5. Apr 28, 2008

zcabral

ok so this is what i did now...
pi [integral 0 to 2 (2x)^2-(2x-x^2)^2]
=
pi*-x^5/5 + 4x^4/4 |[0,2]

6. Apr 29, 2008

dynamicsolo

The "outer radius" of each ring is provided by the curve y = 2x , since that is farther from the x-axis than y = x^2 on the interval [0,2]. So the area of the full disc is (pi)[(2x)^2] . The "hole" is bounded by the curve y = x^2 , so that is the "inner radius" of the ring we want to include in our integration. So the area of the hole in each disc is (pi)[(x^2)^2].

The integral will then be (pi) [integral 0 to 2 of {(2x)^2} - {(x^2)^2} dx]

= (pi) [integral 0 to 2 of (4x^2 - x^4) dx].

7. Apr 29, 2008

tron_2.0

Last edited: Apr 29, 2008