Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 am trying to figure out what is going on in this Surface Area problem: enter image description here

As I attempted to illustrate above, It seems like the formula has been applied incorrectly. Where x has been placed should be f(x) which in this problem is (x^(1/3) + 2).

Is something going on that changes the input of f(x) to x that I am not seeing here? Thanks!

share|cite|improve this question
up vote 1 down vote accepted

We can get a somewhat different point of view by looking at arclength, surface area, and other problems for curves given parametrically.

Let the curve be given by $x=x(t)$, $\:y=y(t)$. Then the surface area obtained when we rotate the chunk of the curve from $t=a$ to $t=b$ around the $x$-axis is $$\int_{t=a}^b 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$ For rotation about the $y$-axis, we have a very similar expression for surface area: $$\int_{t=a}^b 2\pi x\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$

In our case, we have $y=\sqrt[3]{x}+2$. Let's choose a nice parametrization, something that will simplify the calculations. A sensible choice is $x=t^3$, in which case $y=t+2$. Since $x$ goes from $1$ to $8$, $t$ will go from $1$ to $2$.

Calculate. We have $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=1$. For rotation about the $y$-axis, we get surface area $$\int_1^2 2\pi t^3\sqrt{1+9t^4}\,dt.$$ This integral can be calculated by a simple substitution. But this was not your problem.

For rotation about the $x$-axis, we get surface area $$\int_1^2 2\pi (t+2)\sqrt{1+9t^4}\,dt.$$ This is a nightmarish integral. The hardest part, actually the marginally easier $\int\sqrt{1+x^4}\,dx$, has been discussed on this site. It may not be worth looking up.

Whoever was writing out solutions for your book got lucky. If (s)he had not made a mistake, (s)he would have had an awful integral to evaluate. (It would start out looking even worse without the parametric approach.)

share|cite|improve this answer

No, it is incorrect.

It also happens to be example 2 on this page of Paul's Online Math Notes. He explains as best as anyone could hope.

share|cite|improve this answer
Actually paul does the same exact thing in his notes! – Matt Jul 7 '11 at 16:15
@Matt: Paul does a witty revision. Note that in the radical, there is no fraction. – mixedmath Jul 7 '11 at 16:32

The formula you have provided assumes that the surface is being produced by rotating around the x-axis. In that case, f(x) is the radius of rotation. It looks like you might instead be creating the surface by rotating around the y-axis, in which case the radius of rotation would be x, as in the solution. You can think of finding the surface area as a bunch of little slanty sticks each carving out a thin strip of surface as they're carried around the axis of rotation. Each stick (represented by the stuff in the square root) traces out a strip of circumference 2$\pi$r, where r depends on the set-up of the rotation.

Hope that helps from a conceptual standpoint.

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.