Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to find a formula for the surface area of a solid of revolution rotated around the $y$-axis. The curve is $f(x)=x^2$ on $[0,1]$. However, my answer must be in terms of $f$, not $f^{-1}$.

share|cite|improve this question

The only thing that changes when you revolve the curve about the $y$-axis instead of the $x$-axis is the expression for the radius of revolution. When you revolve it about the $x$-axis, the radius of revolution at a particular value of $x$ is $|f(x)|$, the distance from the curve to the $x$-axis; when you revolve it about the $y$-axis, the radius is $|x|$, the distance from the curve to the $y$-axis.

The element $dA$ of surface area at a given $x\in[0,1]$ is still $dA=2\pi x\,ds$, since the radius of revolution about the $y$-axis is simply $x$ in this case. From studying arc length you know that

$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\sqrt{1+\big(f\,'(x)\big)^2}\;,$$ so

$$dA=2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi x\sqrt{1+\left(f\,'(x)\right)^2}\,dx\;;$$

now just integrate $dA$ over the appropriate range of values of $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.