# Calculating the surface area of revolution for parametric equation.

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

$(\frac{ds}{dt})^2=x'^2+y'^2=(t^3-\frac{1}{t})^2$

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

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?

• 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.