# Homework Help: Volume of Solids with Known Cross Sections (Calculus AB)

1. Mar 8, 2010

### Flaneuse

Can anyone help me with this calculus problem? I have tried it a couple of different ways, but keep getting various incorrect answers; I think my problem is probably in the setup.

I am trying to find the volume of the solid created by revolving the area bounded by these two equations
y = 6 - 2x - x$$^{2}$$
y = x + 6
around the x-axis, and around y = 3.

First, I found the intersection points of the two equations, which are at (-3,3) and (0,6)
For the revolution around the x-axis, I tried doing
$$\pi \int ^{0}_{-3}((6 - 2x - x[^2}$$)$$^{2}$$ - (x+6)$$^{2}$$ )dx,

simplified to $$\pi$$ \int ^{6}_{0}(-32x -13x$$^{2}$$ +4x$$^{3}$$ + x$$^{4}$$ )dx ,
and got a final answer of -27$$\pi$$/5
The correct answer is 243$$\pi$$ /5

For the revolution around y = 3, I got -3-x^{2} from subtracting the lower function, x + 6, from the top function, 6 - 2x - x^{2}, and then tried
$$\pi \int^{6}_{0}(6-(-3-x^{2}))^{2} dx, simplified to \pi^{6}_{0}\int (6+3x+x^{2} ) dx, and ended up with 29376[tex]\pi$$/5.

The answer for this one is supposed to be 108$$\pi$$/5

Again, I think my real problems are in the setup, probably with the "top minus bottom" part.

Also, I am new to Physics Forum, and can't seem to get rid of how, in the second integral, my words ended up included in the integral. I would appreciate it if anyone could tell me how to fix/avoid that.

Thanks!

2. Mar 9, 2010

### Staff: Mentor

Note: Fixed up some of your LaTeX by removing a lot of superfluous [ tex] tags.
The first integral above was fine, but there are some mistakes in the second one. You switched the limits of integration. They should still be -3 and 0. Also, the coefficients of the x^4 and x^3 terms are right, but those for the x^2 and x terms are incorrect.
I'll take a look at the 2nd integral in a separate post.

3. Mar 9, 2010

### Staff: Mentor

For the second integral,
$$\Delta V = \pi (R^2 - r^2)\Delta x$$
The limits of integration are the same, namely x = -3 to x = 0. R = the y-value on the parabola - 3, and r = the y-value on the line - 3.

4. Mar 9, 2010

### Flaneuse

Hm, I don't know what I must have been thinking when I decided to start integrating from 0 to 6, especially when I had already started using -3 to 0 as well... I suppose I was particularly jumbled yesterday, or something of that sort.

Thanks so much for the help; goodness knows I needed it. May I ask though, how can I differentiate between [yes, a very bad choice of words; I mean when to use one as opposed to the other, not take the derivative of] when R(x)^2 - r(x)^2 is equal to (top equation minus line)^2 - (bottom equation - line)^2 and when it is equal to (top minus bottom)^2 - (line)^2?

5. Mar 10, 2010

### Staff: Mentor

R is the larger radius and r is the smaller radius. Both radii are as measured from the axis of rotation. If you have drawn a sketch, it should be pretty obvious.

6. Mar 10, 2010

Thanks!