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Triple integrals solution check

  1. May 11, 2015 #1
    Are these correct?
    Thanks in advance!

    1.) Set up the triple integral for ##f(x,y,z) = xy + 2xz## on the region ##0 ≤ x ≤4, 0 ≤ y ≤ 2## and ##0 ≤ x ≤ 3xy + 1##.

    ##\displaystyle \int_0^4 \int_0^2 \int_0^{3xy+1} 2y +2xz\ dz\ dy\ dx##

    [tex]\text{2.) Set up the triple integral in cylindrical coordinates to find the volume bounded by}\\
    z = x^2 + y^2, z = 0, x^2 + y^2 = 1\ \text{and}\ x^2 + y^2 = 4[/tex].

    ##\displaystyle \int_1^2 \int_0^{2\pi} \int_0^{\sqrt{2}} r^2 r\ dr\ d\theta\ dz##
     
  2. jcsd
  3. May 11, 2015 #2

    mfb

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    How did xy become 2y in the first integral?

    I don't understand how you set up the second one, it does not look right.
     
  4. May 11, 2015 #3

    mathman

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    The first one looks funny. Did you mean [itex]0 \le z \leq 3xy+1[/itex] ?
     
  5. May 13, 2015 #4

    HallsofIvy

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    First, yes, in problem 1, you mean "[itex]0\le z\le 3xy+ 1[/itex]". But the integral is set up correctly.

    In problem 2, there is no good reason to write "[itex]r^2r[/itex]" rather than "[itex]r^3[/itex]".
    More importantly, the upper bound on the figure is [itex]z= x^2+ y^2[/itex] which is [itex]z= r^2[/itex] in cylindrical coordinates.

    What you have, [itex]\int_1^2\int_0^{2\pi}\int_0^\sqrt{2} r^3 drd\theta dz[/itex], would be the integral of [itex]r^2[/itex] over the cylinder whose base is a circle with center at (0, 0), radius [itex]\sqrt{2}[/itex], and extending from z= 1 to z= 2. What you want is z to go from 0 go [itex]r^2[/itex] (so the z-integral will have to be inside the r-integral).
     
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