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

[NWeid1]

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 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?

 

[NWeid1]

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

 

[Dick]

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.

 

[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

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

 

[Dick]

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.