Calculating Surface Area of a Revolution Rotated About the Y-Axis

[deerhake.11]

(my first dealings with latex.. so bare with me if this looks a little messed up at first )

Homework Statement

Find the surface area for the equation:
$$x = 3y^{4/3} - \frac{3}{32}y^{2/3}$$

with bounds $$-216 \leq y \leq 216$$

rotated about the Y-axis.

Homework Equations

$$\int^a_b 2\pi f(y) \sqrt{1+(\frac{dx}{dy})^2}$$

The Attempt at a Solution

well... going with that equation i get to this point:

$$2\pi \int^{216}_{-216} (3y^{4/3} - \frac{3}{32}y^{2/3})(4y^{1/3} + \frac{1}{16}y^{-1/3})$$

from there I tried to multiply out the equation and solve the integral with the bounds, but it isn't giving me the correct answer. I'm not sure what I'm doing wrong. I suspect I have to break the integral up smaller pieces but I am not sure where to break it at.

 

[stunner5000pt]

$$\int^a_b 2\pi f(y) \sqrt{1+\left(\frac{dx}{dy}\right)^2}$$

you have substituted dx/dy incorrectly into the equation

dx/dy should be inside the root

 

[HallsofIvy]

Because of the symmetry, you can integrate from 0 to 216 and then multiply by 2. What do you get?

 

[deerhake.11]

ok... when I multiply out the integrand I get:

$$2\pi [12y^{5/3} - \frac{3}{16}y^2 - \frac{9}{2048}y^{1/3}]^{216}_{0}$$

when I evaluate at 216 & 0, i get 7552892.305... I multiply that by 2 (symmetry) and then multiply that by 2pi, which gives me 94925010.28

edit... nevermind, i made a stupid math mistake. Got it right now, thanks a ton HallsofIvy!