Finding Volume of a curve

1. Oct 22, 2009

vipertongn

I'm suppose to rotate around y-axis...

x=8y;x=y^3 and y> and equal to 0

I was told to integrate (x^1/3-0)^2-(1/8x-0)^2

so.... pi integral 0 to sqrt(512) x^2/3-1/64x^2 dx

I end up with 151.65 but that's incorrect can someone tell me what i did wrong and how to get the right answer?

2. Oct 22, 2009

HallsofIvy

Staff Emeritus
Who told you this? This would be the volume of the region rotated around the x- axis. You said you wanted the volume of the region rotated around the y- axis. That would be the integral of $\pi((8y)^2- (y^2^2) dy$

$\sqrt{512}= 16\sqrt{2}$

Last edited: Oct 22, 2009
3. Oct 22, 2009

vipertongn

Oh wait so it should be integral 64y^2-y^6 dy?

when i integrate that way it comes out negative

Last edited: Oct 22, 2009
4. Oct 22, 2009

HallsofIvy

Staff Emeritus
You are right. If forgot the squares (I have edited my response so I can pretend I didn't make thata mistake!). The area of a disk is $\pi r^2$ and you are taking the areas of two disks and subtracting them. The volume you want is $\int_0^{2\sqrt{2}} 64y^2- y^4 dy$.

5. Oct 22, 2009

vipertongn

Ahhh i'm confused ok I started with

x=8y
x=y^3
y>0

the limits of y are 0 to sqrt(512) and the limits of x is 0 to sqrt(8)

SO! Integral should look like 64y^2-y^6 dy Oh gosh nvm THANKS SO MUCH :D umm if i am correct its Rout-Rin right?