# Surface area of x = ln(y) - y^2/8 from 1 to e

1. Mar 4, 2012

### NWeid1

1. The problem statement, all variables and given/known data
Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.

2. Relevant equations
SA = $\int 2\pi*(f(x))*\sqrt{1+[f'(x)]^2}dx$

3. The attempt at a solution
SA = $\int 2\pi*(ln(y) - \frac{y^2}{8}))*\sqrt{1+\frac{9}{16*y^2}}dy$ from 1 to e

I don't know how to solve this integral, which makes me thing I set it up wron.g

2. Mar 4, 2012

### Dick

I don't think you did the sqrt(1+f'(y)^2) part right. Can you show how you worked that out?

3. Mar 4, 2012

### NWeid1

That was the equation that my prof. gave me, lol

4. Mar 4, 2012

### Dick

I'm saying I don't think you simplified 1+f'(y)^2 correctly when f(y)=ln(y)-y^2/8. You should get something inside the square root that becomes a perfect square. I'm asking how you got that part to be 1+9/(16y^2)? I don't think that's right.

5. Mar 4, 2012

### NWeid1

oh ok. I took the derivative of $x = lny - \frac{y^2}{8}$ and got

$x = \frac{1}{y} - \frac{1}{4y}$ and squared it to get

$\frac{1}{y^2} - \frac{1}{2y^2} + \frac{1}{16y^2}$ which is

$\frac{16}{16y^2} - \frac{8}{16y^2} + \frac{1}{16y^2}$ which is

$\frac{9}{16y^2}$

6. Mar 4, 2012

### Dick

The derivative of that is $\frac{1}{y}-\frac{y}{4}$. That's where you are starting to go wrong.