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Multivar Calc. HW problems

  1. Oct 9, 2006 #1
    I got stuck on the 3rd question and now I kinda dunno what to do. Can someone help me a bit ?

    [​IMG]

    I drew a graph and shaded the region W.

    [​IMG]

    I thought it would be a good idea to integrate with respect to Z first. So I got:

    [tex] V = \int \ \int \ \int_{\sqrt{x^2+y^2}}^{\frac{4-x-y}{2}} \,dz[/tex]

    Then I got stuck.
     
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 9, 2006 #2

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    Solve for the curve where the plane intersects the cone in terms of x and y and you'll get the equation of the ellipse in the xy plane that bounds the "shadow" of the shape. Let the integral over x and y go over the interior of this ellipse.
     
  4. Oct 9, 2006 #3
    That's what I have been trying to do but setting
    [tex] \frac{4-x-y}{2} = \sqrt{x^2+y^2}[/tex] leads me to a dead end.

    I get
    [tex]4-x-y-x+\frac{x^2}{4} + \frac{xy}{4} -y+\frac{xy}{4}+\frac{y^2}{4} = x^2 + y^2[/tex]

    [tex]4-2x-2y-2xy = \frac{3}{4}x^2 + \frac{3}{4}y^2[/tex] ?
     
    Last edited: Oct 9, 2006
  5. Oct 9, 2006 #4

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    Try defining new variables u=x+y and v=x-y, which are rotated at 45 degrees to x and y, so that the u and v axes lie along the principal axes of the ellipse.
     
  6. Oct 9, 2006 #5
    I don't see how it helps since I can't define [tex]\sqrt{x^2+y^2}[/tex] in terms of u or v ?
     
  7. Oct 9, 2006 #6

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    Why not? Solve for x and y in terms of u and v.
     
  8. Oct 9, 2006 #7
    Ok so this is what I did.
    x = (u+v)/2 & y = (u-v)/2

    [tex] \frac{4-u}{2} = \sqrt{ \left( \frac{u+v}{2} \right)^2+ \left( \frac{u-v}{2} \right)^2}[/tex]

    Am I on the right path ?

    *edit, yeah u're right. fixed now.
     
    Last edited: Oct 9, 2006
  9. Oct 9, 2006 #8

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    If you used my definition of u and v, you didn't solve for x and y correctly. There shouldn't be a factor of 1/2 on one and not the other.
     
  10. Oct 9, 2006 #9
    So from this I get:
    [tex] \frac{16-8u+u^2}{4} = \frac{2u^2+2v^2}{4} [/tex]

    [tex]16-8u+u^2=2u^2+2v^2[/tex]

    If I go further:

    [tex] 16 = u^2+8u+v^2[/tex]

    [tex] 16 + 16 = u^2+8u+16+v^2[/tex]

    [tex] 32 = (u+4)^2 + v^2[/tex]

    [tex] \frac{(u+4)^2}{32} + \frac{v^2}{16} = 1[/tex]

    Now this is an equation of an ellipse but it is not centered along the z axis. Did I do something wrong ?
     
    Last edited: Oct 9, 2006
  11. Oct 9, 2006 #10

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    That looks right. There's no reason to expect the ellipse to be centered at the z-axis. Imagine the plane was at a much steeper angle, then the ellipse would have one end close to the z-axis and one far away.
     
  12. Oct 9, 2006 #11
    So can I set up the integral this way ?

    [tex] \int_{-4\sqrt{2}}^{4\sqrt{2}} \ \int_{-\sqrt{16- \frac{(u+4)^2}{2}}}^{\sqrt{16- \frac{(u+4)^2}{2}}} \ \int_{\sqrt{ \left( \frac{u+v}{2} \right)^2 + \left( \frac{u-v}{2} \right)^2}}^{\frac{4-u}{2}} \,dz\,dv\,du
    [/tex]

    P.S. Plz let this be right. Editing the code for this is a *****. ! :p
     
    Last edited: Oct 9, 2006
  13. Oct 9, 2006 #12

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    I'm not gonna check all your algebra, but that looks like the right idea. Just one more thing: when you change variables from x,y to u,v you need to remember that dxdy is not just replaced by dudv, there is an extra factor. This is the jacobian determinant, or if you're not familar with this, you can derive it geometrically by seeing the area of the parellelogram in u,v space corresponding to the tiny rectangle dx by dy.
     
  14. Oct 9, 2006 #13
    Thank you for your help. I should be fine after this.:smile:
     
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