Question

Calculus help with integration and finding volumes?
A solid lies between the planes perpendicular to the x-axis at x = sqrt(3) and x = -sqrt(3) The cross-sections perpendicular to the x-axis are circles whose diameters stretch from y=4/sqrt(1+x^2) to y=-4/sqrt(1+x^2) Find the volume of the solid in cubic units.

Step by step instructions would be wonderfully helpful!

Answer 1

OK. Let's walk through this.

Answer 2

Sure! I believe this would be the disk method, so you would integrate piR^2. I am not sure about what to do with the + and - y equations as R, they are the same except for their signs. I also do not understand how I would integrate (4/sqrt(1+x^2))^2 - I know that if it was not squared this would fit the arctan(x) integration rule, but it is squared so I am not sure what to do.

Answer 3

So I would just integrate \[4\div(\sqrt{1+x ^{2}}\] then? I'm still not entirely sure how to do this.

Answer 4

Sorry, I re-read the question, you're right.

Answer 5

Is this for single-variable or multi-variable? For single you do need to do it with cross-sections, for multivar you can do it with a double or triple integral.

Answer 6

I think single? I haven't heard of double or triple integrals. Also, if this helps, the answer is 32pi^2/3. I just would like to know how to get there.

Answer 7

Yeah, this is for single. So we're going to add up a lot of circles to get our shape.

Answer 8

We know that the radius of each circle is \[\frac{4}{\sqrt{1+x^2}}\]so the area of each circle is \[\frac{16\pi}{1+x^2}\]right?

Answer 9

Okay yes I see how that would happen.

Answer 10

So now we're just going to integrate that from -\sqrt{3} to \sqrt{3} (this is basically the disk method, but you're really only supposed to do that with solids of rotation).

Answer 11

\[\int^\sqrt{3}_{-\sqrt{3}} \frac{16\pi}{1+x^2} dx=16\pi \int^\sqrt{3}_{-\sqrt{3}} \frac{dx}{1+x^2} =16\pi\left[ \arctan(x) \right]^\sqrt{3}_{-\sqrt{3}}=16\pi\left(\frac{\pi}{3}-\frac{-\pi}{3}\right)=\frac{32\pi}{3}
\]Sorry for the delay, my LaTeX skills are subpar.

Answer 12

Okay, that looks pretty good except for the fact that pi is supposed to be squared. Where would that happen?

Answer 13

It is. To continue that line which got cut off,
\[16\pi\left(\frac{\pi}{3}-\frac{-\pi}{3}\right)=16\pi\frac{2\pi}{3}=\frac{32\pi^2}{3}\]

Answer 14

Thank you!