Surface Area Revolving Around Y-Axis: Confused?

[vipertongn]

Homework Statement

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.

Homework Equations

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

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...

 

[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...

 

[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

 

[lanedance]

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

 

[lanedance]

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

 

[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