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Setting up a Double Integral in Polar Coordinates

  1. Oct 18, 2016 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Consider the 'ice cream cone' bounded by
    z = sqrt1a.gif 14 − x2 − y2 and z = sqrt1a.gif x2 + y2
    .(a) Find the equation of the intersection of the two surfaces in terms of x and y.
    (b) Set up the integral in polar coordinates.

    2. Relevant equations


    3. The attempt at a solution

    I got part a without any trouble. You just set each equation equal to one another and get:
    ##x^2+y^2=7##

    I also found my limits of integration just fine: ##\int_0^{2\pi} \int_0^{\sqrt{7}} rdrd\theta##

    I just can't seem to set up the integrand correctly. I thought it would be: ##x^2+y^2-7 = (rcos(\theta))^2+(rsin(\theta))^2 - 7 = r^2 (cos^2(\theta)+sin^2(\theta)) - 7 = r^2-7##

    That would make the integral: ##\int_0^{2\pi} \int_0^{\sqrt{7}} (r^3-7r) drd\theta ##

    However that's not correct and I can't seem to find the correct one.
     
  2. jcsd
  3. Oct 18, 2016 #2

    SammyS

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    I suppose those equations should read

    ##z = \sqrt{ 14 − x^2 − y^2\ }\ ## and ##\ z = \sqrt{ x^2 + y^2\ }## .

    What is it that integrating is supposed to give you ?
     
  4. Oct 18, 2016 #3

    Drakkith

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    Whoops, I meant to fix those before I posted...

    Some sort of ice cream cone shape. Perhaps I'm supposed to integrate using those other two equations instead of where they intersect?
     
  5. Oct 18, 2016 #4

    LCKurtz

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    You need to integrate$$
    \iint_R z_{upper} - z_{lower}~dA$$where ##R## is the region you have figured out, and the ##z##'s are expressed in terms of ##r,\theta##.
     
  6. Oct 19, 2016 #5

    SammyS

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    My question could have been more specific.

    Are you integrating to find surface area, or to find volume?
     
  7. Oct 19, 2016 #6

    Drakkith

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    I'm pretty sure it's volume. We haven't done anything with surface area yet.

    Ah, of course. It's just like Calc 2 where you find the region between two lines, except this time it's 2 surfaces.

    So that should be:
    ##\int_0^{2\pi}\int_0^\sqrt{7} (\sqrt{14-r^2} - r) rdrd\theta = \int_0^{2\pi}\int_0^\sqrt{7} (r\sqrt{14-r^2} - r^2) drd\theta##
    Because ##x^2+y^2 = r^2##, so ##\sqrt{14-x^2-y^2} = \sqrt{14-(x^2+y^2)} = \sqrt{14-r^2}##
    And ##\sqrt{x^2+y^2} = \sqrt{r^2} = r##

    Now to integrate...
     
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