MHB Josh Mcdaniel's question at Yahoo Answers regarding a volume of revolution

[MarkFL]

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.

 

[MarkFL]

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:

$$dV=\pi r^2\,dx$$

where:

$$r=y=3x^4$$

Hence, we have:

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

Summing up the disks, we find:

$$V=2\cdot9\pi\int_0^1 x^8\,dx$$

Applying the FTOC, we obtain:

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

Shell method:

The volume of an arbitrary shell is:

$$dV=2\pi rh\,dy$$

where:

$$r=y$$

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

Hence, we find:

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

And so, summing all the shells, we find:

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

Application of the FTOC yields:

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