# Homework Help: Integration Problems

1. Feb 9, 2005

### Sombra

if I want to find the volume of y= (x + 2) ^(1/2) about the x-axis with the boundaries of -1 and 1, then would it be :

V = pi times integral of (x+2) dx?? Its confusing trying to set it up. The volume of a section would be v = pi r^2 (delta x)

Also I'm having trouble finding the volume of the region enclosed by y = (2-x)(2+x), y = 3x, and the y- axis when rotated around the x - axis. It's hard to even picture what the figure will be like, much harder to find the volume. Please point me in the right direction! Thanks!!

2. Feb 9, 2005

### NateTG

You seem to have the first part, but don't forget the limits of integration.
$$\pi\int_{-1}^{1}(x+2) dx$$

For the second part, one thing that jumped out at me was:
$$y=(2-x)(2+x)=2^2-x^2=4-x^2$$
is a parabola that opens downward, so it will hit the line
$$y=3x$$ in two (or zero) places.

So, let's look for intersections by solving:
Subtract
$$y=3x$$
from
$$y=x^2-4$$
to get
$$0=x^2-3x-4$$
so we can use the quadratic formula
$$x=\frac{3\pm\sqrt{9-4(1)(-4)}}{2}=\frac{3 \pm 5}{2}$$

That tells us that the functions intersect where $x=-1$ and where $x=4$. Those will be the limits of integration. A quick check, say at $x=0$ confirms that $4-x^2 > 3x$ in the region.

So, to calculate the volume, you can (equivalently) calculate the volume from $$y=4-x^2$$
on and then subtract the volume from
$$y=3x$$
on the region.
Or you can integrate using annuli (annuluses?) instead of disks.

3. Feb 9, 2005

### Sombra

So, for the first one, the answer would be 4pi, correct?

For the second one, I thought the 2 functions interesected at (1,3), so I said the volume was pi times integral (with boundaries of 0 and 1) of [(4-x^2) -(3x)] dx since it has to be subracted. But I think I need to find the integral of r^2 times delta x, so would i have to square (4-x^2) and make it (16-x^4) then subtract 3x (or would it be 9x^2)??? Thanks