General

Are they part of Calc II?

@BadCereal: Like I said it doesn't really matter. But do you really want to take the derivative with respect to x?
Then it'll turn out like this: 2pi Integral (x*sqrt(1+(1/(3x^(2/3))))
Yea...that's really ugly to solve.
@NonSonoFronz: Wait till you get to Taylor and Maclaurin series. God those were brutal imo </3

[quote=hyperfire7]Give me a sec, I'll work it out on paper.

Edit: Ok just so you know you're completely clear, the formula in finding the surface area on a cartesian revolution is this:
S.A. = Integral of the following: (2*pi*(x or y)) * (sqrt(1+(dy/dx)^2)) dx or dy
Unlike traditional integrals, dx or dy is upto your preference. Some problems will be easier when taking the derivative in respect to X, and vice versa. This problem looks easier with respect to Y.

So make y = cuberoot(x) into x = y^3.
So dx/dy = 3y^2.
@NonSonoFronz
1) S.A. = Integral (2*pi*y^3) * (sqrt(1+(3y^2)^2)) dy
2) Take out the 2pi since it's a constant: 2pi * Integral (y^3 * (sqrt(1+(9y^4)) dy)
3) Use u-sub: u = 1+9y^4
du = 36y^3 dy
du/36 = y^3 dy

4) Set the new bounds since we used u-sub.
Ua = 1+9(1)^4 = 10
Ub = 1+9(2)^4 = 145

5) And the rest should be straight from Calc I.
2pi * (1/36) Integral (sqrt(u)), 10 < u < 145.
You should be able to solve from here. Answer should be Pi/27 * (145*sqrt(145) - 10*sqrt(10))[/quote]

Thanks a bunch for your input.
One thing, how did you get y^3 in step 1? Everything else makes sense.
Nvm, @NonSonoFronz: cleared it up.

Thanks a lot you two!

Since we're rotating it about the y-axis, we need the function in terms of x.
So x = y^3

There is a formula for calculating the surface area which is S = INT(2pi*x*ds) while ds = sqrt(1+(dx/dy)^2)dy)
So you need to find the derivative of x which is 3y^2. Now you can just plug and chug. ds = sqrt((1+9y^4)dy)
Then you can use u-sub. u = 1+9y^4 du = 36y^3 du/36 = y^3
Now ds = sqrt(u)du

Just reset your boundaries in terms of u and you should be golden.

Someone let me know if I did this wrong, because I'm attempting to teach myself calc II at the moment.

Oh man. Looks like I'm on the right track. <3

Give me a sec, I'll work it out on paper.

Edit: Ok just so you know you're completely clear, the formula in finding the surface area on a cartesian revolution is this:
S.A. = Integral of the following: (2*pi*(x or y)) * (sqrt(1+(dy/dx)^2)) dx or dy
Unlike traditional integrals, dx or dy is upto your preference. Some problems will be easier when taking the derivative in respect to X, and vice versa. This problem looks easier with respect to Y.

So make y = cuberoot(x) into x = y^3.
So dx/dy = 3y^2.

1) S.A. = Integral (2*pi*y^3) * (sqrt(1+(3y^2)^2)) dy
2) Take out the 2pi since it's a constant: 2pi * Integral (y^3 * (sqrt(1+(9y^4)) dy)
3) Use u-sub: u = 1+9y^4
du = 36y^3 dy
du/36 = y^3 dy

4) Set the new bounds since we used u-sub.
Ua = 1+9(1)^4 = 10
Ub = 1+9(2)^4 = 145

5) And the rest should be straight from Calc I.
2pi * (1/36) Integral (sqrt(u)), 10 < u < 145.
You should be able to solve from here. [b]Answer should be Pi/27 * (145*sqrt(145) - 10*sqrt(10))[/b] or whatever the hell that is in decimal.