Triple integral volume bounds

  • Thread starter ProPatto16
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  • #1
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Homework Statement



evaluate triple integral of z.dV where the solid E is bounded by the cylinder y2+z2=9 and the planes x=0 and y=3x and z=0 in the first octant

Homework Equations



for cylindrical polar co-ords, x=rcos[tex]\theta[/tex], y=rsin[tex]\theta[/tex] and z=z

The Attempt at a Solution



im just struggling to grasp the bounds here. the cylinder has x as its centre line. and r=3. which means shape extends out from x 3 units along y and z axis's. and extends along x from origin 3x units. then stops due to plane on y. thats about as much as i can gather. the projected region that i should take the volume of the solid over should be projected onto the yz plane for this case. which would show a quarter circle with r = 3 right? with y=sqrt(9-z2) with y>0 so achieve first quadrant.
but i cant actually work out what to integrate each integral between.

help? much appreciated!
 

Answers and Replies

  • #2
HallsofIvy
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The given cylinder, [itex]y^2+ z^2= 1[/itex] has axis on the x-axis and its curved side projects to a circle in the yz plane. I would d you "swap" x and z in setting up the cylindrical coordinates:
[itex]x= x[/itex]
[itex]y= r sin(\theta)[/itex]
[itex]z= r cos(\theta)[/itex]

Now the limits of integration should be easy.
 
  • #3
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Yeah I knew that bit. R=3 and theta=pi/2 for first octant also. So looking at quarter circle is first quadrant of yz plane. Y goes from 0-3 and z goes from 0-3 and therefore with x=y/3 then x goes from 0 to 1?
 
  • #4
326
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so then the integral needing evaulation is.. i cant do latex so ub means upper bounds and lb means lower bound

[tex]\int[/tex](ub 3, lb 0)[tex]\int[/tex](ub 3, lb 0)[tex]\int[/tex](ub 1, lb 0) z.dz.dy.dx??
 

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