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Double (volume) integral

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data
    - Sketch the region V of 3-space that is bounded above and below by the two surfaces z=Z1(x,y) = 0 and z=Z2(x,y)=1+x^2+y^2 and where the domain of these functions is the region R in the xy plane enclosed by the four lines y=x, y=-x, y=2+x and y=2-x

    -Calculate the volume using a double integral namely

    [​IMG]
    3. The attempt at a solution

    For the first part I have firstly drawn the four lines in the xy plane
    [​IMG]

    Obviously I make it 3d by adding the top (i.e. 1+x^2+y^2 for -1 ≤ x ≤1 and 0 ≤ y ≤ 2)

    My question is regarding the volume integral. It is difficult to nominate the y bounds due to the nature of the 4 lines. I am thinking I should split the area into two sections, bisected by the y axis such that the volume can be calculated by:

    [​IMG]

    Am I on the right track?
     
    Last edited: Oct 6, 2008
  2. jcsd
  3. Oct 6, 2008 #2

    gabbagabbahey

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    Looks like a good plan to me!:approve:
     
  4. Oct 6, 2008 #3
    Thanks for the quick reply

    hmmm, noticed that I need to reverse the order of integration (noting also I must change the terminals) though as before I was ending up with x's after the second integration. That is, I need to perform integration with respect to y then with respect to x such that

    [​IMG]
     
  5. Oct 6, 2008 #4

    gabbagabbahey

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    Yes, I thought your original order of integration was a typo.
     
  6. Oct 6, 2008 #5
    also noticed my terminals are wrong!

    for the first part with respect to y I should be integrating from x to -x+2 and for the second part I should be integrating from -x to x+2.

    After fixing it up I got a volume of 14/3 units cubed. Is there any way to check this answer?
     
    Last edited: Oct 6, 2008
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