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Triple integral question

  • #1
I= [∫(0 to 1) ∫(0 to 2z) ∫(z to 1) dx dy dz] + [ ∫(0 to 1) ∫(2z to 1+z) ∫(y-z to 1) dx dy dz]

1) evaluate
2) use order dy dx dz, along with the new bounds

my attempt for 1) got me an answer of -7/6

for part 2) i'm having trouble getting the correct bounds. the bounds from my attempt are
D1 = {(x, y)|0 ≤ x ≤ 2z, z ≤ y ≤ 1}
D2 = {(x, y)|2z ≤ x ≤ 1+z, x+z ≤ y ≤ 1 }

i can do integration easily, and am nearly finished with this question, but its the bounds that are prohibiting me from progressing any further.

please any help would be good. i tried trying 2d graphs and so forth already...
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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In the first integral, z ranges from 0 to 1; for each z, y ranges from 0 to 2z; for each (y,z), x ranges from z to 1.

Since your new "order", dydxdz, still has z as the "outer" integral, and the limits of integration for x and y depended only on z, there is no reason to change the limits at all:
[tex]\int_{z=0}^1\int_{x= z}^1\int_{y= 0}^{2z} dy dx dz[/tex]

For the second integral, z ranges from 0 to 1 again, y ranges from 2z to 1+ z, and x ranges from y- z to 1. Since z is still the "outer" integral, this is really just swapping x and y and we can treat z as a constant. x= y- z is the same as y= x+ z which crosses the vertical lines y= 2z and y= z+ 1 at x= z and x= 1 respectively. That is, the middle, "x", integral will be from 1 to z. The three lines, y= x+ z, x=2z, and x= 1 form a trapezoid with base y= 0 and top y= x+ z. The integral is
[tex]\int_{z=0}^1\int_{x= 1}^z\int_{y= 0}^{x+ z} dy dx dz[/tex].
 

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