Find the volume of the solid obtained by rotating the region bounded by $y = x$ and $x = y ^ { 3 }$ about the $x$-axis.
Select the correct answer.
A) $16 \pi$
B) $\frac { 7 } { 2 }$
C) $\frac { 4 } { 15 } \pi$
D) $\frac { 18 } { 35 }$
E) $\frac { 16 } { 7 } \pi$

Question

Quizplus · Accepted Answer

First, we find the intersection point of the two curves:

$$y=x$$
$$x=y^3$$

Substituting for $x$ in the first equation,

$$y=y^3$$
$$\Rightarrow y(y-1)(y+1) = 0$$

So, $y = -1,0,1$. Since we are rotating about the x-axis, the volume we want consists of three solids that are generated by rotating the parts of the region which are above the x-axis; one for $y \in [0,1]$, one on $y \in [-1,0)$, and one for $y = 0$, respectively. For each part, we use the washer method. (See picture.)

[asy]

import solids;

size(150);

revolution c=cone(1,0.5),d=c;

draw(surface(c),gray(0.8),nolight);
draw(surface(d),gray(0.6),nolight);

draw(c,black+lw(0.6));
draw(d,black+lw(0.6));

draw((0,0,0)--(1.1,0,0),Arrow3(6));
draw((0,-0.6,0)--(0,0.8,0),Arrow3(6));
draw((0,0,-0.8)--(0,0,0.8),Arrow3(6));

draw(shift(0,0,0)*yscale(2)*yscale(0.05)*zscale(0.05)*"$y$");
draw(shift((1.2,0,0))*xscale(0.5)*yscale(0.5)*zscale(0.05)*"$x$");

draw((0,0)--(1,1.1),red,Arrow(size=0.2cm),EndArrow);
draw((0,0)--(1/1.44,-1/1.44),red,Arrow(size=0.2cm),EndArrow);
draw((0,1)--(1,1),red,Arrow(size=0.2cm),EndArrow);

label("$y=x$",(0.8,1.3,0));
label("$x=y^3$",(0.8,-0.8,0));

[/asy]

Consider the part where $y$ is between $0$ and $1$. Then we get a volume of

$$\pi \int_0^1 (y^3)^2 - y^2 dy = \pi \int_0^1 y^8 - y^2 dy = \frac{4}{15}\pi.$$

By symmetry, the part over $[-1,0)$ also contributes $\frac{4}{15}\pi$, so the total volume up to this point is $\frac{8}{15}\pi$. Finally, for the part where $y=0$, we just have a disk with radius 0, so the volume of this part is simply 0. Therefore, the total volume is $\boxed{\frac{4}{15}\pi}.$

Find the Volume of the Solid Obtained by Rotating the Region