Surface area of a revolution

• Mar 26th 2011, 06:53 PM
alexgeek
Surface area of a revolution
I'm having some trouble with a question, I've got a lot of the difficult stuff down but I can't get the limits right and my answer is twice what it should be.
Quote:

Curve C: $x=cos^3 \theta \: , \: y= sin^3 \theta \: , \: 0 \leq \theta \leq \frac{\pi}{2}$

The curve C is rotated through 360 degrees about the x-axis. Show that the curved surface area of the solid of revolution generated by:
$
6 \pi \int^{ \frac{\pi}{2} }{0} }_0 sin^4 \theta cos \theta d\theta$

Hence find this curved surface area.
So I know that $A = \int 2 \pi y ds$ and that $ds = \sqrt{ (\frac{dx}{d \theta})^2 + (\frac{dy}{d \theta})^2 } = \frac{3}{2}sin 2 \theta$.

So:

$A = 2\pi \int 2 sin^3 \theta \frac{3}{2} \sin 2 \theta d \theta$
$sin 2 \theta \equiv 2 \sin \theta \cos \theta$
$A = 6\pi \int sin^4 \theta cos \theta d \theta$

So I have the right integral but I can't show that the limits are 0 to $\frac{\pi}{2}$.

I don't think it wants me to use symmetry as that would make it $12 \pi$, I tried $\theta = arcos( x^{\frac{1}{3} } )$ but I can't get the right answers.

Cheers
• Mar 26th 2011, 06:55 PM
Prove It
Surely it's because you're told in the first line of the problem what the values of $\displaystyle \theta$ are...
• Mar 26th 2011, 06:59 PM
alexgeek
Well it says show, and I can't show that they are the limits.
I tried the lower limit as -1:
$\theta = arcos( (-1)^{\frac{1}{3} }) = \pi$
And upper as 1:
$\theta = arcos( 1^{\frac{1}{3} }) = 0$
So I'm quite confused.
• Mar 26th 2011, 07:02 PM
Prove It
Quote:

Originally Posted by alexgeek
Well it says show, and I can't show that they are the limits.
I tried the lower limit as -1:
$\theta = arcos( (-1)^{\frac{1}{3} }) = \pi$
And upper as 1:
$\theta = arcos( 1^{\frac{1}{3} }) = 0$
So I'm quite confused.

No, it says to show that the integral gives you the surface area. You don't need to go any further than to use the formula you have been given (which you have done) and substituting the endpoints of your $\displaystyle \theta$ domain.
• Mar 26th 2011, 07:19 PM
alexgeek
To be picky, it actually says show that the integral between those two points gives the surface area.
Could you perhaps, live up to your name of "prove it" and show me why the limits have to be what it says, as to me, mathematically they can't be.
Thanks
• Mar 26th 2011, 07:37 PM
Prove It
I have just noticed that you have made a mistake in your integral. It appears you took out $\displaystyle 2\pi$ as a factor, but also left $\displaystyle 2$ inside the integral as well. That will explain why you're getting double the answer you're supposed to.

It always helps to have a picture of what you are trying to do. In this case, to draw a graph of the function will mean you need to write $\displaystyle y$ in terms of $\displaystyle x$, and the required range of $\displaystyle \theta$ will be the distance between $\displaystyle x$ intercepts.

If $\displaystyle x = \cos^3{\theta}$ and $\displaystyle y = \sin^3{\theta}$, then

$\displaystyle y = \left(\sin^2{\theta}\right)^{\frac{3}{2}}$

$\displaystyle y = \left(1 - \cos^2{\theta}\right)^{\frac{3}{2}}$

$\displaystyle y = \left[1 - \left(\cos^3{\theta}\right)^{\frac{2}{3}}\right]^{\frac{3}{2}}$

$\displaystyle y = \left(1 - x^{\frac{2}{3}}\right)^{\frac{3}{2}}$.

The $\displaystyle x$ intercepts are where $\displaystyle y = 0$, so

$\displaystyle 0 = \left(1 - x^{\frac{2}{3}}\right)^{\frac{3}{2}}$

$\displaystyle 0 = 1 - x^{\frac{2}{3}}$

$\displaystyle x^{\frac{2}{3}} = 1$

$\displaystyle x^2 = 1$

$\displaystyle x = \pm 1$.

And since $\displaystyle x = \cos^3{\theta}$, solving $\displaystyle \cos^3{\theta} = \pm 1$ will give $\displaystyle \theta = 0$ and $\displaystyle \theta = \pi$, but you can now use symmetry, since we have found why your answer is double what it should be.
• Mar 26th 2011, 07:43 PM
alexgeek
Why do I do these things?! Thanks for that :)
I see what I need to do now, and I wouldn't have thought to equate y to zero in order to get the upper and lower limits so thanks for that too.
• Mar 27th 2011, 02:47 PM
alexgeek
Sorry I've gone over it again and I made a typo in my original post but the answer is still what I said it was:

$A= 2\pi \int \sin^3 \theta \frac{3}{2}\sin 2 \theta d\theta$
$A =3\pi \int \sin^3 \theta 2 \sin \theta \cos \theta d \theta$
$A =6\pi \int \sin^4 \theta \cos \theta d \theta$

So lower limit should be $b = arcos(-1) = \pi$
And upper limit $a = arcos(1) = 0$

So shouldn't it be: $A =6\pi \int^{\pi}_0 \sin^4 \theta \cos \theta d \theta$

So still, using symmetry would make it $12 \pi$ which is wrong.

I keep checking it and can't see where I'm wrong :S