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Did I set up my integral correctly?

  1. Nov 12, 2015 #1
    1. The problem statement, all variables and given/known data
    use triple integral to find volume of the solid in the first octant that is bounded above by[tex] x+2y+3z=6[/tex] and laterally by the clyinder [tex]x^2+y^2=4[/tex]

    2. Relevant equations

    3. The attempt at a solution
    From the given plane I got:
    [tex]Z=2-\frac{1} {3}x-\frac{2} {3}y[/tex]
    from the given Cylinder i got:

    since its only interested in the first octant, lower bounds should all be zero.

    so I get [tex]\int_{x=0}^2\int_{y=0}^\sqrt{4-x^2} \int_{z=0}^{2-\frac{1} {3}x-\frac{2} {3}y} \,dz \, dy \, dx[/tex]

    my concern is with Z's lower bound, I think it should be 0 as well, but if using [tex]z=x^2+y^2-4[/tex] seems like also okay, but being this way, this cylinder became a different solid, it became an infinite paraboloid. so that means I cant just set [tex]z=x^2+y^2-4[/tex] ? because the Z value of an infinite cylinder does not depend on x or y?
  2. jcsd
  3. Nov 12, 2015 #2


    Staff: Mentor

    Looks fine to me.
    Yes. You're in the first octant, where ##x \ge 0, y \ge 0##, and ##z \ge 0##.
    No, it's not. The equation of the cylinder is ##x^2 + y^2 = 4##. z is completely arbitrary, so you can't just stick it in the equation.
  4. Nov 12, 2015 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The "equation" ##z = x^2+y^2-4## has no connection to the problem, because the lowest point on the plane and the curve ##x^2+y^2=4## has positive ##z##. Therefore, the lowest value of ##z## is ##z = 0## and the top face of the cylinder does not come down that far. In short, you setup is correct.

    BTW: thanks for using a typed version, which in this case is do-able because no diagrams are needed.
  5. Nov 12, 2015 #4
    got it. thank you all.
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