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Evaluate the triple integral

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  1. Oct 21, 2015 #1
    Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid
    x = 5y2+ 5z2 and the plane x = 5.

    My work so far:
    Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y2+5z2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Also, when x=0, y2 + z2=0 -----> r=0 and x=5 -----> y2+z2=1 -----> r=1, so my bounds for r are 0 ≤r ≤ 1.

    My triple integral is then:
    ∫∫∫E 5x dV = ∫(from 0 to 2π) ∫(from 0 to 1) ∫(from 5r2 to 5) 5xr dx dr dθ

    I keep getting the wrong answer after evaluating it
     
  2. jcsd
  3. Oct 21, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What answer do you get when you evaluate it? What do you think is the correct answer?
     
  4. Oct 22, 2015 #3
    Your domain of integration is ##E=\{(x,y,z) \in\mathbb{R}^3, \ 0\le x \le 5, 0\le y^2 + z^2 \le x/5\}##

    If you sketch what it looks like, you'll see that a natural change of variable is ##(x,y,z) = \phi(x',\rho,\theta) = (x', \rho \cos\theta, \rho \sin\theta) ##, with ##\rho > 0## and ##\theta \in [0,2\pi[## to make it bijective except for the points on the ##x## axis.

    Then, your reciprocal domain is ##\phi^{-1}(E) = \{ 0\le x' \le 5,\ 0<\rho\le \sqrt{x'/5}, \ 0\le \theta < 2\pi \} ##
     
  5. Oct 22, 2015 #4

    Mark44

    Staff: Mentor

    Please don't delete the three parts of the homework template. Its use is required.
     
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