# Find the volume of the solid formed by rotating the region

• zcabral
In summary, the student attempted to solve a homework problem using the "disk method", but realized that the region is not (pi)(2x - [x^2])^2 , but rather the area of the outer circle minus the area of the inner circle. This caused incorrect results, as the integral is not given by the expression he is using.

## Homework Statement

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^2
y=2x

## Homework Equations

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

## 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?

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

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.

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

zcabral said:
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

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?

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]

zcabral said:
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]

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].

took a stab at it. i think yr on the right track, this was more so to see if i could still do it lawl.

http://i21.photobucket.com/albums/b277/riceboy89/problem.jpg [Broken]

(im going to d/l latex)

Last edited by a moderator: