# Volumes: The Disk Method

1. Nov 9, 2008

### elitespart

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 $$\pi$$x$$^{2}$$. How do I find the limits of intergration and put x in terms of H and R? Thanks.

Last edited: Nov 9, 2008
2. Nov 9, 2008

### Staff: Mentor

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.

3. Nov 9, 2008

### elitespart

So x^2 = y-H(-R^2/H) and then run that from 0 to H?

4. Nov 9, 2008

anybody?

5. Nov 9, 2008

### gabbagabbahey

sounds good, what do you get when you integrate that?

6. Nov 9, 2008

### elitespart

I feel like an idiot but I'm having trouble integrating this. Can someone get me started on it? Thanks.

7. Nov 9, 2008

### gabbagabbahey

$$\int_0^H \pi x^2 dy= \int_0^H \pi (y-H)(\frac{-R^2}{H}) dy =\frac{\pi R^2}{H} \int_0^H (H-y)dy$$

....surely, you know how to integrate this?

8. Nov 9, 2008

### elitespart

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.