Center of mass and integrating

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Homework Statement


A solid "rough" of constant density ([tex]\delta[/tex] =1) is bounded below by the surface z=4y^2,above by the plane z=4, and one the ends by the planes x=1 and x=-1. Find the center of mass...


Homework Equations


[tex]_{M}[/tex]xy =[tex]\int\int\int[/tex] z [tex]\delta[/tex] dV
Mass = [tex]\int\int\int[/tex][tex]\delta[/tex]dV
then to find the center of mass, the z component would be [tex]_{M}[/tex]xy / M

The Attempt at a Solution


[tex]_{M}[/tex]xy = [tex]\int\int\int[/tex][tex]^{4}_{4y^2}[/tex] z [tex]\delta[/tex]dzdydx
=[tex]\int\int[/tex] [z^2/2][tex]^{4}_{4y^2}[/tex]dydx
=[tex]\int[/tex][tex]^{1}_{-1}[/tex][tex]\int[/tex][tex]^{1}_{0}[/tex](8-8y^4)dydx
=8[tex]\int^{1}_{-1}[/tex][y-y^5/5][tex]^{1}_{0}[/tex] dx =8[tex]\int[/tex][tex]^{1}_{-1}[/tex][4/5]dx
=8[4x/5][tex]^{1}_{-1}[/tex] = 8[4/5+4/5]= 64/5

For some reason, and it isn't obvious to me why, the solution manual says that this answer should be 128/5. i'm off by a factor of 2, but i don't know what i did wrong. I haven't attempted to find the mass yet, but i think once i understand what I did wrong here, that answer shouldn't be hard to find.
 

Answers and Replies

  • #2
HallsofIvy
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Why are you integrating, with respect to y, from y= 0 to y= 1? The "cylinder" z= 4y2 intersects the plane z= 4 at y= -1 and 1.
 
  • #3
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Why are you integrating, with respect to y, from y= 0 to y= 1? The "cylinder" z= 4y2 intersects the plane z= 4 at y= -1 and 1.
i don't know why that didn't register with me. thanks for pointing that out. my answer is now 128/5, the same as the book's.
 

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