# Thread: Volumes: The Disk Method

1. ## Volumes: The Disk Method

1. If the area bounded by the parabola y = H - (H/R^2)x^2 and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolution with height H and radius of base R. Show its volume is half the volume of the circumscribing cylinder

Okay so the thickness of the disk is dy and the area is (pi)x^2. How do I find the limits of integration and put x in terms of H and R? (assuming that is the right path to take) Thanks.

2. Originally Posted by elitespart
1. If the area bounded by the parabola y = H - (H/R^2)x^2 and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolition with height H and radius of base R. Show its volume is half the volume of the circumscribing cylinder

Okay so the thickness of the disk is dy and the area is x^2. How do I find the limits of integration and put x in terms of H and R? (assuming that is the right path to take) Thanks.
you need to fix the whole physcisforum.com thing, and not by double posting.

anyway, do you know what the disc method says?

note, you need to write the integral in terms of y, that is, you need to integrate dy. thus, find the y-intercepts and that will guide you as to how to find the limits. if you want to keep things in terms of x, you need to use the shell method, not the disk

(also, did you draw a diagram? those always help. plus, have you already posted this on physicsforums?)

3. Originally Posted by Jhevon
you need to fix the whole physcisforum.com thing, and not by double posting.

anyway, do you know what the disc method says?

note, you need to write the integral in terms of y, that is, you need to integrate dy. thus, find the y-intercepts and that will guide you as to how to find the limits. if you want to keep things in terms of x, you need to use the shell method, not the disk

(also, did you draw a diagram? those always help. plus, have you already posted this on physicsforums?)
fixed it. Yeah I posted it there a while ago. No reply so I put it here. So do I integrate from 0 to H with the integrand being y - H(R^2/H)(pi)?

4. Originally Posted by elitespart
fixed it. Yeah I posted it there a while ago. No reply so I put it here. So do I integrate from 0 to H with the integrand being y - H(R^2/H)(pi)?
ok, for the limits. but note that if $y = H - \frac {H}{R^2}x^2$ then $x = \pm \sqrt{\frac {R^2(H - y)}H}$

next time, please state that you posted the problem on another forum explicitly, and post the link while you're at it. that way, we don't waste time answering questions that might have been answered elsewhere. where it not for that quirk when posting, we would not have known

5. hm but since radius is x wouldn't the integral be (pi)x^2 dy and it'd be replaced with (R^2(H-y))/H?

And thanks for letting me know. I'll post the link next time.