# Thread: volume of a solid

1. ## volume of a solid

i'll give the problem then explain my issue with it.
the solid lies between planes perpendicular to the x axis at x=-1 and x=1. The cross sections perp. to the x axis are circular disks whose diameters run from the parabola y=x^2 to the parabola y=2-x^2

seems pretty simple but for the life of me i can't figure it out. I figured to use the washer method, with y=2-x^2 being R and y=x^2 being r. so i set that up and integrate it, and get volume = pi(4x-2x^2/3) . (There were two other functions, but one was positive, one was negative, so they cancel out, leaving this). I need to evaulate that from -1 to 1. but doing that results in 0, which obviously can't be the answer. where did i go wrong?

2. You are revolving about the x-axis from x=1- to x=1

Using washers:

${\pi}\int_{-1}^{1}[x^{4}-(2-x^{2})^{2}]dx$

Which simplifies to $4{\pi}\int_{-1}^{1}(1-x^{2})dx$

3. ??? I see no reference in the problem to "rotating" anything around the x-axis.

Imagine dividing the solid into thin sheets of thickness " $\Delta x$". Each sheet can be approximated by a disk with upper boundary at $y= 2- x^2$ and lower boundary at $y= x^2$ for some x in that section of the x axis. Thus, each disk has diameter $2- x^2- x^2= 2- 2x^2$ and so radius $1- x^2$. The area of each disk is $\pi(1- x^2)^2= \pi(1- 2x^2+ x^4)$ and so the volume is $\pi(1- 2x^2+ x^4)\Delta x$. The approximate area, the sum of those disks, is $\sum\pi(1- 2x^2+ x^4)\Delta x$. That "Riemann sum", in the limit, becomes the exact area $\pi\int_{x= -1}^1 (1- 2x^2+ x^4)dx$.

The integral galactus gives is correct for the region between $y= 2- x^2$ and $y= x^2$, rotated around the x-axis but that does not appear to be the problem as stated: "The cross sections perp. to the x axis are circular disks whose diameters run from the parabola y=x^2 to the parabola y=2-x^2". I would certainly not call the line between inner and outer boundaries of a washer a "diameter".

4. Oh, I see. I quickly assumed it was a S.O.R. I see now what you mean, HOI. Sorry 'bout that. I was at the wheel

5. Sorry about the confusion, my mistake.

Thanks very much for all your help, both of you. Much appreciated