I solved a problem using a method that's completely different from the mark scheme and I got the right answer, but I'm unsure whether or not it's just some coincidence. Here's the question:

The curve $C$ has parametric equations $$ x=t^2, y=\frac{1}{4}t^4-\ln{t}$$ for $1\leq t\leq 2$. Find the are of the surface generated when $C$ is rotated $2\pi$ radians about the y-axis.

Here's what I did ( not very rigorous):

I thought cutting the curve with horizontal cuts should do the trick. So I just evaluated $2\pi\int _{t=1}^2xdy$ since $dy=(t^3-\frac{1}{t})dt$ it's just $$2\pi\int_1^2(t^5-t)dt$$ which is fairly easy to find.

Here's what's in the mark scheme:

$x'=2t$ and $y'=t^3-\frac{1}{t}$


$S=\int 2\pi x ds=2\pi\int_1^2(t^5-t)dt$

My understanding is that both answers slice $C$ differently and so maybe this was just some coincidence. I can't prove that both methods are equivalent though. So if someone can give me an idea as to why they are that'd be great. Also, it seems to me that the method I used was way easier, is there a place where it's better to use that other method?



  • Your proposed formula is, unfortunately, incorrect. A more detailed formulation of the question occurs in Surface Area of Solid of Revolution Derivation. This currently has no answer, but a comment refers to Doubt in Application of Integration - Calculation of volumes and surface areas of solids of revolution, which has excellent answers (one detailed, one elegantly concise).

  • If you examine the mark scheme's computation, you'll find a difference of sign with what's in your post: $x' = 2t$ and $y' = (t^{3} - \frac{1}{t})$, so $$ \left(\frac{ds}{dt}\right)^{2} = x'(t)^{2} + y'(t)^{2} = \left(t^{3} + \tfrac{1}{t}\right)^{2}, $$ n.b., not $\left(t^{3} - \tfrac{1}{t}\right)^{2}$, which would be equal to $y'(t)^{2}$ itself. If the mark scheme really has a minus sign on the right, that's an error.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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