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