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.