YAC(alc)Thread - integration to solve for volume

[Fenixgoon]

ok, i got the equations X = Y^2 and X=4. im supposed to find the volume of the solid of revolution about the line X = 6.

what i did: r1 = Y^2, r2 = 2 (difference between 6 and 4)

(pi) integral from of (R1)^2 - (R2)^2

give you Pi (integral) Y^4 - 4

Pi [( Y^5 divided by 5) - 4y]

the answer is 384(Pi)/5

two things confuse me: not sure from what points to take integral (done 0-4, 0-2, 0-6) and not sure what else is making me wrong.

 

[GigaCluster]

Are you saying that 384(Pi)/5 is what you get, or what the correct answer is?

 

[Fenixgoon]

Originally posted by: GigaCluster
Are you saying that 384(Pi)/5 is what you get, or what the correct answer is?

384(Pi)/5 is the correct answer.. but i don't understand how the book gets that (i dont come anywhere close)

 

[GigaCluster]

Hmm. I am working on it.

 

[GigaCluster]

The best I can get is 17.16*pi. I used the Disk Method with a vertical axis of rotation. Are you sure that you reproduced the problem accurately?

 

[Fenixgoon]

about as best as i can do.. i'll just ask my teacher tmw

 

[jagec]

I got the book answer.

 

[GigaCluster]

Please enlighten us!

 

[jagec]

OK, first I transferred the problem so the axis of rotation is the Y-axis (x=0). So, you have what is basically a torus with the inside at x=2 and the outside at x=(6-y^2). So, you just take the area of the outer circle pi*r^2, minus the area of the inner circle pi*r^2, times dy, and integrate from y=-2 to y=2.

int((pi*(6-y^2)^2-pi*2^2)dy,y,-2,2) = pi*384/5

pic

 

[oog]

subtract the integral of the "hole" of the disc from the integral of the outer volume, pretending like there is no hole.