Men's question at Yahoo Answers regarding a volume of revolution

[MarkFL]

Here is the question:

Find the volume of the resulting solid by any method?

y^2-x^2=1, y=2 about y-axis.

thank you in advance.

Here is a link to the question:

Find the volume of the resulting solid by any method? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

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.