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Finding the volume between two surfaces

  1. Jul 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume between z = x^2 + y^2 and z = 3 - x - y


    2. Relevant equations
    None

    3. The attempt at a solution
    I must use a double integral. Using polar coordinates I find that the volume is equal to:

    V = ∫∫(3 - rcosθ - rsinθ) r dr dθ - ∫∫r^2 r dr dθ

    I'm struggling trying to find the region of integration.
    I found that the projection onto the x-y plane is the circle x^2 + x + y^2 + y = 3, or (x+1/2)^2+(y+1/2)^2 = 7/2. By switching to polar coordinates I get: r^2 + rcos + rsinθ = 3

    Since it's a circle, I assume that 0 ≤ θ ≤ 2∏.
    But what about r?

    I tought It went from 0 <= r <= √(7/2), but I'm not sure anymore since it's a circle centered outside the origin. What can I do?
     
    Last edited: Jul 5, 2012
  2. jcsd
  3. Jul 5, 2012 #2

    LCKurtz

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    Those limits would be correct if the circle was centered at the origin, and ##r## and ##\theta## were measured from there. Your integral is $$
    \iint_R 3-x-y-(x^2+y^2)dydx =\iint_R 3-(x+y+x^2+y^2)dydx=\iint_R\frac 7 2 - (x+\frac 1 2)^2-(y+\frac 1 2)^2 dydx$$Now if you let ##x+\frac 1 2 = r\cos\theta## and ##y+\frac 1 2 = r\sin\theta## and ##dydx=rdrd\theta## your limits would be correct. This centers your polar coordinates at the center of the circle.
     
  4. Jul 5, 2012 #3
    Let me see if I understand. What you did was translating both surfaces and the projection to (-1/2, -1/2)?
     
  5. Jul 6, 2012 #4

    HallsofIvy

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    Rephrase that please. It doesn't make sense to talk about moving an entire surface to a single point. Do you mean translate the surfaces so that (-1/2, -1/2) is translated to (0, 0)?
     
  6. Jul 6, 2012 #5
    Yes, sorry. That was what I wanted to say. To make (-1/2, -1/2) the new origin. That's what happened?
     
  7. Jul 6, 2012 #6

    LCKurtz

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    I guess you could think of it that way. I prefer to think of it as putting the polar coordinates at the center of the circle. Let ##(x,y)## be any point inside the circle. The position vector ##\vec R=\langle x,y\rangle## goes from the origin to that point. Let ##\vec R_0=\langle -\frac 1 2,-\frac 1 2\rangle## be the vector from the origin to the center of the circle. Now if you let ##r,\theta## be the polar coordinates from the center of the circle to ##(x,y)## then the vector from the center of the circle to ##(x,y)## is ##\vec R_1=\langle r\cos\theta,r\sin\theta\rangle##. Now you have ##\vec R =\vec R_0+\vec R_1## or ##
    \langle x,y\rangle=\langle -\frac 1 2,-\frac 1 2\rangle+\langle r\cos\theta,r\sin\theta\rangle##.
     
  8. Jul 7, 2012 #7
    Now I get it. Thank you so much :)
     
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