This one is going to be a little difficult to explain without being able to draw out pictures but hopefully you can make sense of it.
The lower boundary of your region is x^2.
Solve x=y^2 for y to find our upper boundary.
The upper boundary turns out to be the positive root, sqrt(x).
Your curves intersect at 0 and 1.
So that gives us our boundaries for integrating.
This part is rather difficult to explain in text but I'll try...
If you slice out a "thin" rectangle in the x-direction of your region,
and imagine spinning it around y=-5,
you'll end up with a washer (a "thin" disk with a hole in it).
The outer radius is the length from -5 to our upper boundary:
(sqrt(x) + 5)
The inner radius is the length from -5 to our lower boundary:
(x^2 + 5)
To find the volume of our washer (which we'll call dV),
we'll multiply the surface area: pi*(R^2-r^2)
by the "thickness" of the washer: dx
[tex]\rm dV=\pi[R^2-r^2]dx[/tex]
[tex]\rm dV=\pi[(\sqrt x+5)^2-(x^2+5)^2]dx[/tex]
Expand out the squares and combine like-terms,
[tex]\rm dV=\pi[-x^4-10x^2+x+10\sqrt x~]dx[/tex]
To find our total volume,
we sum up all of these individual washer volumes,
[tex]\rm V=\int dv[/tex]
[tex]\rm V=\int_0^1 \pi[-x^4-10x^2+x+10\sqrt x~]dx[/tex]
Pull the pi out front,
apply your power rule for integration to each term,
plug in your boundary values and do all the subtraction.
You should end up with something like: 109pi/30
which is approximately 11.414
I don't think I made any boo boos along the way... but it's certainly possible.
Lemme know if that's not clear.
