Volume of revolution for 3 over x cubed

Key concept: This is a standard disk-method volume problem. The radius of each cross-section is the function value $f(x)=\frac{3}{x^3}$, so the volume is found by integrating the area of a circle along the interval.

Step by step

Set up the volume integral

When a curve is rotated about the $x$-axis, the disk method gives

$V(t)=\pi\int_1^t [f(x)]^2\,dx.$

Here,

$f(x)=\frac{3}{x^3},$

so

$V(t)=\pi\int_1^t \left(\frac{3}{x^3}\right)^2 dx=\pi\int_1^t \frac{9}{x^6}\,dx.$

Compute the antiderivative

Rewrite the integrand as a power:

$\frac{9}{x^6}=9x^{-6}.$

Then

$\int 9x^{-6}\,dx=9\cdot \frac{x^{-5}}{-5}=-\frac{9}{5}x^{-5}.$

So

$V(t)=\pi\left[-\frac{9}{5}x^{-5}\right]_1^t.$

Evaluate the definite integral

Substitute the bounds:

$V(t)=\pi\left(-\frac{9}{5t^5}+\frac{9}{5}\right).$

Factor the constant:

$V(t)=\frac{9\pi}{5}\left(1-\frac{1}{t^5}\right).$

So the volume as a function of $t$ is

$\boxed{V(t)=\frac{9\pi}{5}\left(1-\frac{1}{t^5}\right)}.$

Why this works

The radius of each disk is the distance from the curve to the $x$-axis, and the area of a disk is $\pi r^2$. Since the rotation is around the axis, the square of the function value appears inside the integral. The variable upper limit $t$ makes the result a function, not just a single number.

Pitfall alert

A frequent mistake is to integrate $f(x)$ instead of $[f(x)]^2$. For volumes of revolution around the $x$-axis, the disk method uses area, so the square is essential. Another common error is forgetting that the lower limit is fixed at 1, which changes the final expression into a function of $t$. Also, because the curve is positive on $[1,t]$ for $t>1$, there is no need for absolute values here.

Try different conditions

If the solid were generated by rotating the same curve around the $y$-axis instead of the $x$-axis, the shell method would be more appropriate and the formula would change completely. If the interval were $[a,t]$ rather than $[1,t]$, the same setup would give $V(t)=\pi\int_a^t 9x^{-6}dx=\frac{9\pi}{5}\left(a^{-5}-t^{-5}\right)$. The method is the same, but the lower bound alters the constant term.

Further reading

disk method, solid of revolution, definite integral