# Homework Help: Area of a rotated surface

1. Feb 27, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data
Find the exact value when $y = \frac{x^3}{2} + \frac{1}{6x}$ in domain $\frac{1}{2} \leq x \leq 1$ is rotated around the X axis

2. Relevant equations
Area is given by
$$A = \int \pi \left(f(x)\right)^2$$

3. The attempt at a solution
If the above formula is correct, then I should be integrating the following

$$A = \int_{\frac{1}{2}}^{1} ( \frac{x^3}{2} + \frac{1}{6x})^2 dx$$

Is this set up alone correct?

2. Feb 27, 2013

### CompuChip

I think the formula you are using is for the volume of the solid of rotation: you can cut it up into thin slices of surface area $\pi r^2 = \pi f(x)^2$ and "thickness" dx.

Do you have a formula with a $2\pi$ and an f'? :)

3. Feb 27, 2013

### stunner5000pt

Hmm i think I am mistaken...
I was trying to look this up and foudn the following formula:

$$A = \int_{a}^{b} 2 \pi f(x) \sqrt{1 + (f'(x))^2} dx$$

is this the fomula I need to be using?

Last edited: Feb 27, 2013
4. Feb 27, 2013

### Dick

If it's surface area you want, then that looks right.

5. Feb 27, 2013

### stunner5000pt

Ok thats perfect...

Now just to clarify, the integral becomes:

$$2\pi \int_{\frac{1}{2}}^{1} \left( \frac{3x^4+1}{6x^2} \right) \sqrt{1+\left(\frac{9x^4-1}{6x^2}\right)^2} dx$$

So I was a bit lazy so I plugged into wolfram alpha and got the answer $445\pi/768$

is this correct? is there anything that I missed to consider?

6. Feb 27, 2013

### Dick

Conceptually you aren't missing anything. In detail, it might be wrong. [Edit: my mistake, it is correct].

Last edited: Feb 27, 2013
7. Feb 27, 2013

### stunner5000pt

Thank you :)
Been stressing about these questions for quite some time now