Surface area of a revolution

  • #1
deerhake.11
7
0
(my first dealings with latex.. so bare with me if this looks a little messed up at first :rolleyes: )

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 im not sure where to break it at.
 
Last edited:

Answers and Replies

  • #2
stunner5000pt
1,447
2
[tex]\int^a_b 2\pi f(y) \sqrt{1+\left(\frac{dx}{dy}\right)^2}[/tex]

you have substituted dx/dy incorrectly into the equation

dx/dy should be inside the root
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
43,021
970
Because of the symmetry, you can integrate from 0 to 216 and then multiply by 2. What do you get?
 
  • #4
deerhake.11
7
0
ok... when I multiply out the integrand I get:

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

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!
 
Last edited:

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