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men's question at Yahoo! Answers regarding a volume of revolution

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MarkFL

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Feb 24, 2012
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MarkFL

Administrator
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Feb 24, 2012
13,775
Hello men,

The first thing we should do is plot the region to be revolved:

men.jpg

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:

\(\displaystyle dV=\pi r^2\,dy\)

where:

\(\displaystyle r=x=\sqrt{y^2-1}\)

and so we have:

\(\displaystyle dV=\pi (y^2-1)\,dy\)

Summing the disks by integrating, we find:

\(\displaystyle V=\pi\int_1^2 y^2-1\,dy=\pi\left[\frac{1}{3}y^3-y \right]_1^2=\)

\(\displaystyle \pi\left(\left(\frac{1}{3}2^3-2 \right)-\left(\frac{1}{3}1^3-1 \right) \right)=\pi\left(\frac{8}{3}-2-\frac{1}{3}+1 \right)=\frac{4\pi}{3}\)

Shell method:

Substituting for $y$ into the hyperbolic equation, we find the positive root:

\(\displaystyle 4-x^2=1\)

\(\displaystyle x^2=3\)

\(\displaystyle x=\sqrt{3}\)

So, we will integrate from $0\le x\le\sqrt{3}$.

The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=2-\sqrt{x^2+1}\)

and so we have:

\(\displaystyle dV=2\pi (2x-x\sqrt{x^2+1})\,dx\)

Summing the shells by integration, we find:

\(\displaystyle V=2\pi\int_0^{\sqrt{3}} 2x-x\sqrt{x^2+1}\,dx=2\pi\left[x^2-\frac{1}{3}(x^2+1)^{\frac{3}{2}} \right]_0^{\sqrt{3}}=\)

\(\displaystyle 2\pi\left(\left(3-\frac{1}{3}(3+1)^{\frac{3}{2}} \right)-\left(0-\frac{1}{3}(0+1)^{\frac{3}{2}} \right) \right)=2\pi\left(3-\frac{8}{3}+\frac{1}{3} \right)=\frac{4\pi}{3}\)

To men and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.