Center of mass and integrating

[flame_m13]

Homework Statement

A solid "rough" of constant density (\delta =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

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

The Attempt at a Solution

_{M}xy = \int\int\int^{4}_{4y^2} z \deltadzdydx
=\int\int [z^2/2]^{4}_{4y^2}dydx
=\int^{1}_{-1}\int^{1}_{0}(8-8y^4)dydx
=8\int^{1}_{-1}[y-y^5/5]^{1}_{0} dx =8\int^{1}_{-1}[4/5]dx
=8[4x/5]^{1}_{-1} = 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.

 

[HallsofIvy]

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.

 

[flame_m13]

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.