Triple Integral ( IS THIS RIGHT?)

  • Thread starter een
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  • #1
een
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Triple Integral ( IS THIS RIGHT??)

Homework Statement


Let R be the solid enclosed by the planes x=0, y=0, z=2 and the surface x^2+y^2, where
x[tex]\geq[/tex]0, y[tex]\geq[/tex]0. Compute[tex]\int\int\int[/tex]xdxdydz

Homework Equations





The Attempt at a Solution


I did [tex]\int[/tex]0-1[tex]\int[/tex]0-1[tex]\int[/tex](x^2+y^2)-2 xdzdydx

[tex]\int[/tex]0-1[tex]\int[/tex]0-12x-x^3-xy^2 dydx
[tex]\int[/tex]0-1[2xy-x^3y-xy^3/3]0-1 dx
[tex]\int[/tex]0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
Can someone tell me if this is the correct answer
 

Answers and Replies

  • #2
lanedance
Homework Helper
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pretty hard to read, below might help your formatting...
[tex]\int_{0}^1dx\int_{0}^1dy[/tex]
 
  • #3
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you got x and y's upper limits wrong.
you are calculating the volume between these two surfaces (z(x,y) = x2 + y2, z < 2), see attachment.
 

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  • #4
een
11
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okay.. this is what i got...
the lower limit of x is 0 the upper limit is sqrt(2)
same for y
the lower limit of z is x^2+y^2 and the upper limit is 2...
is that right?
 
  • #5
113
0


yes, my result is 4/3.
 
  • #6
een
11
0


i am getting sqrt(2)/3... what am i doing wrong?... i used the limits that you told me and im pretty sure that i integrated correctly... can you show me step by step
 
  • #7
35,124
6,866


I have switched the order of integration, leaving the integrand unchanged. I think these are the correct limits of integration.
[tex]\int_{x = 0}^{\sqrt{2}} \int_{y = 0}^{\sqrt{2 - x^2}} \int_{z = x^2 + y^2}^{2} x~ dz dy dx [/tex]

The projection of this region onto the x-y plane is the disk x^2 + y^2 <= 2, a disk of radius sqrt(2) centered at the origin. Geometrically, we have a stack of blocks dx by dy that extend from the paraboloid up to the plane z = 2. Then we let the stack sweep out along the y-axis from y = 0 to y = sqrt(2 - x^2) (i.e., out to the boundary of the disk), then finally let this wall of blocks sweep from x = 0 to x = sqrt(2).
 

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