Hello men,
The first thing we should do is plot the region to be revolved:
https://www.physicsforums.com/attachments/860._xfImport
Because of the symmetry across the axis of revolution (both functions are even), we need only concern ourselves with the shaded first quadrant region.
As a means of checking our work, we may use both the disk and shell methods.
Disk method:
We see that when $x=0$ then $y=1$, and so we will integrate from $1\le y\le2$.
The volume of an arbitrary disk is:
$$dV=\pi r^2\,dy$$
where:
$$r=x=\sqrt{y^2-1}$$
and so we have:
$$dV=\pi (y^2-1)\,dy$$
Summing the disks by integrating, we find:
$$V=\pi\int_1^2 y^2-1\,dy=\pi\left[\frac{1}{3}y^3-y \right]_1^2=$$
$$\pi\left(\left(\frac{1}{3}2^3-2 \right)-\left(\frac{1}{3}1^3-1 \right) \right)=\pi\left(\frac{8}{3}-2-\frac{1}{3}+1 \right)=\frac{4\pi}{3}$$
Shell method:
Substituting for $y$ into the hyperbolic equation, we find the positive root:
$$4-x^2=1$$
$$x^2=3$$
$$x=\sqrt{3}$$
So, we will integrate from $0\le x\le\sqrt{3}$.
The volume of an arbitrary shell is:
$$dV=2\pi rh\,dx$$
where:
$$r=x$$
$$h=2-\sqrt{x^2+1}$$
and so we have:
$$dV=2\pi (2x-x\sqrt{x^2+1})\,dx$$
Summing the shells by integration, we find:
$$V=2\pi\int_0^{\sqrt{3}} 2x-x\sqrt{x^2+1}\,dx=2\pi\left[x^2-\frac{1}{3}(x^2+1)^{\frac{3}{2}} \right]_0^{\sqrt{3}}=$$
$$2\pi\left(\left(3-\frac{1}{3}(3+1)^{\frac{3}{2}} \right)-\left(0-\frac{1}{3}(0+1)^{\frac{3}{2}} \right) \right)=2\pi\left(3-\frac{8}{3}+\frac{1}{3} \right)=\frac{4\pi}{3}$$
To men and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.
