Finding Volume Limits and Integration for Paraboloid of Revolution

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The discussion focuses on finding the volume of a solid formed by revolving a parabola around the y-axis, specifically the parabola defined by y = H - (H/R²)x². The volume is shown to be half that of the circumscribing cylinder with height H and base radius R. Participants discuss the use of the disk method for integration, with the incremental volume expressed as πx²Δy. The integration limits are established from y = 0 to y = H, and the relationship between x and y is clarified to facilitate the integration process. Ultimately, the integration confirms the expected volume result, resolving the initial query.
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Volumes: The Disk Method [Resolved]

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 \pix^{2}. How do I find the limits of intergration and put x in terms of H and R? Thanks.
 
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Your disks run from y = 0 to y = H.
The incremental volume of each disk is \pi x^2 \Delta y. For any point (x, y) on the parabola, you have a relationship between x and y, so use this relationship to replace x^2.
 
So x^2 = y-H(-R^2/H) and then run that from 0 to H?
 
anybody?
 
elitespart said:
So x^2 = y-H(-R^2/H) and then run that from 0 to H?

sounds good, what do you get when you integrate that?
 
I feel like an idiot but I'm having trouble integrating this. Can someone get me started on it? Thanks.
 
\int_0^H \pi x^2 dy= \int_0^H \pi (y-H)(\frac{-R^2}{H}) dy =\frac{\pi R^2}{H} \int_0^H (H-y)dy

...surely, you know how to integrate this?
 
wooow. there's something wrong with me. So I ended up with the volume of a cylinder over 2. Which is exactly what I needed. Thank you very much.
 

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