Finding the Volume of a Rotated Curve: Solving for the Correct Integral Limits

[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?

 

[HallsofIvy]

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

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

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

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

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?

 

[vipertongn]

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

when i integrate that way it comes out negative

 

[HallsofIvy]

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.

 

[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?