Discrepancy in calculating the volume formed by revolving the region above $y=x^3$ under $y=1$ and between $x=0$ and $x=1$ about the $x$-axis

Question

Here's the problem:

Find the volume formed by revolving the region above $y=x^3$ under $y=1$ and between $x=0$ and $x=1$ about the $x$-axis

Now here's the integral I came up with:

$\int_{0}^{1}\pi\left(x^{3}-1\right)^{2}dx=2.02$, since the radius is $x^3-1$.

But when I plug it in Wolfram and others, I get $0.45$. What's wrong?

Answer 1

You slices are not disks: they are annuli (disks with holes). The outer radius of the annulus at $x$ is $1$, and the inner radius is $x^3$, so you should be computing

$$\pi\int_0^1\left(1^2-(x^3)^2\right)\,dx=\pi\int_0^1(1-x^6)\,dx=\pi\left[x-\frac{x^7}7\right]_0^1=\frac{6\pi}7\;.$$

The volume that you computed in Wolfram is that generated by the region between the $x$-axis and the curve $y=x^3$ from $x=0$ to $x=1$, not the volume generated by the region between $y=x^3$ and $y=1$.

Answer 2

For each $x\in [0,1]$ the revolving segment is $[x^3,1]$ which yields an annulus of area $\pi(1^2-(x^3)^2)$. Therefore the volume should be $$\int_{x=0}^{1}\pi(1^2-(x^3)^2)\,dx=\frac{6\pi}{7}\approx 2.6928.$$

Answer 3

The volume of revolution is bounded between the curves $y_1=1$ & $y_2=x^3$ from $x=0$ to $x=1$. Consider a disk of revolution of differential volume $\pi(y_1^2-y_2^2)\ dx$ & integrate it with proper limits to find the volume of revolution about x-axis as follows $$\int\pi(y_1^2-y_2^2)dx=\int_0^1\pi(1^2-(x^3)^2)dx=\frac{6\pi}{7}\ \mathrm{unit}^3$$