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

## Homework Statement

Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.

## Homework Equations

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

## 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

Dick
Homework Helper

## Homework Statement

Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.

## Homework Equations

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

## 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

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

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

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

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.

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

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

Dick
Homework Helper
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

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

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