1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Iterated Integral Surface Area Problem (with Polor Coordinates)

  1. Mar 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the surface area of the part of the cone z = sqrt[(x^2 + y^2)] lying inside the cylinder x^2 - 2x + y^2 = 0.



    2. The attempt at a solution

    Partial Derivative x = x/sqrt(x^2 + y^2)
    Partial Derivative y = y/sqrt(x^2 + y^2)

    so...
    sqrt((Partial Derivative Y) ^2 + (Partial Derivative at Y) ^ 2 + 1) =

    sqrt[(x^2 + y^2)/(x^2 + y^2) + 1] = sqrt[2]

    so...
    the surface area =
    integral on E of sqrt[2] r dr d(theta),
    where E is the region (r, theta)|(0 < r < 2, 0 < theta < pi)

    My books solution has the body of the integral as 5, and E as (0<r<1, 0<theta<pi)

    I'm pretty sure I've just made an algebraical mistake in the body, but for the limits I'm confused. I don't know how to treat a cylinder that isn't centered at the origin. I thought r was always the distance from the origin... is it really the distance from the center of the cylinder?

    Thanks for any help or advice you can give me.
     
  2. jcsd
  3. Apr 1, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That depends on where you put the origin of your coordinate system. x2- 2x+ y2= 0 is the same as (x-1)2+ y2= 1, a circle of radius 1 with center at (1, 0). In "straight" polar coordinates, that circle is given by [itex]r= 2cos(\theta)[/itex]. To integrate over that, let [itex]\theta[/itex] go from 0 to [itex]\pi[/itex] (0 to [itex]2\pi[/itex] would go around the circle twice), r from 0 to [itex]2cos(\theta)[/itex].

    Simpler would be to use the transformation [/itex]x= 1+ rcos(/theta)[/itex], [itex]y= r sin(\theta)[/itex], shifting the origin of the coordinate system. Then integrate with [itex]\theta[/itex[/itex] going from 0 to [itex]2\pi[/itex], r from 0 to 1.

    Of course, since the integrand is a constant ([itex]\sqrt{2}[/itex], I don't see how your book can have 5), the integral is just that constant times the area of the circle, [itex]\pi\sqrt{2}[/itex].
     
  4. Apr 1, 2007 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That depends on where you put the origin of your coordinate system. x2- 2x+ y2= 0 is the same as
    (x-1)2+ y2= 1, a circle of radius 1 with center at (1, 0). In "straight" polar coordinates, that circle is given by [itex]r= 2cos(\theta)[/itex]. To integrate over that, let [itex]\theta[/itex] go from 0 to [itex]\pi[/itex] (0 to [itex]2\pi[/itex] would go around the circle twice), r from 0 to [itex]2cos(\theta)[/itex].

    Simpler would be to use the transformation [itex]x= 1+ rcos(/theta)[/itex], [itex]y= r sin(\theta)[/itex], shifting the origin of the coordinate system. Then integrate with [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex], r from 0 to 1.

    Of course, since the integrand is a constant ([itex]\sqrt{2}[/itex], I don't see how your book can have 5), the integral is just that constant times the area of the circle, [itex]\pi\sqrt{2}[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Iterated Integral Surface Area Problem (with Polor Coordinates)
Loading...