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Josh Mcdaniel'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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Here is the question:

Revolving a region about the x axis and finding the volume?


Find the volume of the solid generated by revolving the region bounded by the x axis, the curve y=3x^4 and lines x=-1 and x=1 about the x axis.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Josh Mcdaniel,

Because the region to be revolved is symmetric across the $y$-axis, we need only consider the first quadrant part of the region, and then double the result.

Disk method:

The volume of an arbitrary disk is:

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

where:

\(\displaystyle r=y=3x^4\)

Hence, we have:

\(\displaystyle dV=\pi \left(3x^4 \right)^2\,dx=9\pi x^8\,dx\)

Summing up the disks, we find:

\(\displaystyle V=2\cdot9\pi\int_0^1 x^8\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle V=2\pi\left[x^9 \right]_0^1=2\pi\left(1^9-0^9 \right)=2\pi\)

Shell method:

The volume of an arbitrary shell is:

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

where:

\(\displaystyle r=y\)

\(\displaystyle h=1-x=1-\left(\frac{y}{3} \right)^{\frac{1}{4}}\)

Hence, we find:

\(\displaystyle dV=2\pi y\left(1-\left(\frac{y}{3} \right)^{\frac{1}{4}} \right)\,dy=2\pi\left(y-\frac{1}{\sqrt[4]{3}}y^{\frac{5}{4}} \right)\,dy\)

And so, summing all the shells, we find:

\(\displaystyle V=2\cdot2\pi\int_0^3 y-\frac{1}{\sqrt[4]{3}}y^{\frac{5}{4}}\,dy\)

Application of the FTOC yields:

\(\displaystyle V=4\pi\left[\frac{1}{2}y^2-\frac{4}{3^{\frac{9}{4}}}y^{\frac{9}{4}} \right]_0^3=4\pi\left(\left(\frac{9}{2}-4 \right)-0 \right)=4\pi\cdot\frac{1}{2}=2\pi\)