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Volume bounded by 3 surfaces, did I do this correctly?

  1. Aug 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid bounded by the surfaces
    ## (x^2 + y^2 + y)^2 = x^2 + y^2 ##
    ##x + y + z = 3 ##
    and ##z = 0##

    2. Relevant equations


    3. The attempt at a solution
    I begin by converting to polar coordinates to do a cylindrical integration with 3 variables.

    ## (x^2 + y^2 + y)^2 = x^2 + y^2 ## becomes ##r^4 + 2r^3\sin\theta + r^2\sin\theta = r^2##

    ##x + y + z = 3 ## becomes ##z = 3 - r\cos\theta - r\sin\theta##

    I graphed the first equation and found that 2 is the maximum value of r. Now I write a triple integral in cylindrical coordinates.

    $$V = \iiint_E f(r\cos\theta,r\sin\theta,z) r\,dz\,dr\,d\theta$$

    then
    $$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{3-r\cos\theta-r\sin\theta}r\,dzdrd\theta$$
    then
    $$\int_{0}^{2\pi}\int_{0}^{2}3r-r^2\cos\theta-r^2\sin\theta\,drd\theta$$
    then
    $$\int_{0}^{2\pi}6-\frac{8}{3} \cos\theta-\frac{8}{3} \sin\theta\,d\theta$$
    which then becomes
    $$[12\pi + \frac{8}{3}] - [-\frac{8}{3}]$$
    final answer
    $$12\pi + \frac{16}{3}$$

    I just want to know if what I did makes sense and if this is the correct answer.

    Thanks in advance!
     
  2. jcsd
  3. Aug 2, 2015 #2
    You need to take into account that r depends on theta, and as such you can't simply integrate it from 0 to 2 like you would with a cylinder of constant radius.

    After a bit of algebra you should find that r is bounded between 0 and 1-sinθ, so the integral you want to evaluate is:

    [tex] \int_0^{2\pi} \int _0^{1-sin\theta} \int_0^{3-rcos\theta-rsin\theta} rdzdrd\theta[/tex]


    which I found to be 23/4 pi, but you may want to double check that for yourself.
     
  4. Aug 2, 2015 #3
    Ah, thank you so much!! I knew it felt too easy ?:)
     
  5. Aug 2, 2015 #4
    Can you help me understand how you found that r is bounded by 1-sinθ and 0?
     
  6. Aug 2, 2015 #5
    So you have (x^2+y^2+y)^2 = x^2+y^2.

    After converting to cylindrical coordinate and a bit of rearranging you have
    [tex] r^2 + r(sin\theta-1) = 0 [/tex]

    Solving for r using the quadratic formula will give you r = 0 and r = 1-sinθ.
     
  7. Aug 3, 2015 #6

    LCKurtz

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    Surely, you would just factor out an ##r## giving ##r(r+(\sin\theta-1))=0## instead of using the quadratic formula.

    But that isn't the whole story. What you actually get upon changing to polar coordinates is$$
    r^4 + 2r^3\sin\theta+\sin^2\theta = r^2$$Upon dividing out ##r^2## and losing the trivial ##r=0##:$$
    r^2+2r\sin\theta + \sin^2\theta = 1$$ $$
    (r+\sin\theta)^2-1=0$$ $$
    (r+\sin\theta + 1)(r+\sin\theta -1) = 0$$So you actually get two solutions: ##r = \pm 1 - \sin\theta##. It turns out they both give the same cardioid but traced differently.
     
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