- #1

deerhake.11

- 7

- 0

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

Find the surface area for the equation:

[tex]x = 3y^{4/3} - \frac{3}{32}y^{2/3}[/tex]

with bounds [tex]-216 \leq y \leq 216[/tex]

rotated about the Y-axis.

[tex]\int^a_b 2\pi f(y) \sqrt{1+(\frac{dx}{dy})^2}[/tex]

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

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

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.

## Homework Statement

Find the surface area for the equation:

[tex]x = 3y^{4/3} - \frac{3}{32}y^{2/3}[/tex]

with bounds [tex]-216 \leq y \leq 216[/tex]

rotated about the Y-axis.

## Homework Equations

[tex]\int^a_b 2\pi f(y) \sqrt{1+(\frac{dx}{dy})^2}[/tex]

## The Attempt at a Solution

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

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

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.

Last edited: