# Surface Area

1. Feb 23, 2009

### vipertongn

1. The problem statement, all variables and given/known data

I apologize for the mass questions. I am very most confused with this one

surface area generated by revolving around y axis the curve y=cuberoot(x) from y= 1 to 2.

2. Relevant equations

S 2pi*g(y) $$\sqrt{1-(g'(y))^2}$$

3. The attempt at a solution

i found that g(y)=y3
g'(y)=3y^2

so (g'(y))2=9y4

however, I'm lost at what else to do next...

2. Feb 23, 2009

### lanedance

hmmm.. as they're circles rotated, imagine unrolling a strip it will have area dA = 2.pi.r.ds where r is the raidius of the circle

figure out the radius & look at setting up an integral...

Last edited: Feb 23, 2009
3. Feb 23, 2009

### vipertongn

OH i forgot I was able to get to here

S 2piy3*$$\sqrt{1-9y^4}$$

from there i don't know how to take the antiderivative

4. Feb 23, 2009

### lanedance

ok sorry, how how change of variable to u = 1-9y^4

5. Feb 23, 2009

### lanedance

ok sorry, how how change of variable to u = 1-9y^4 ...

6. Feb 23, 2009

### vipertongn

ok then i end up getting -pi/18(2/3u^3/2) -->-pi/18(2/31-9y^4^3/2)

yea i end up with an incorrect answer...it should end up as pi/27(145sqrt(145)-10sqrt(10)

never mind i set up the problem wrong i got it now ^^ thanks so much

Last edited: Feb 23, 2009