Centroid of a Solid (triple integral)

[Knissp]

Homework Statement

Find the centroid of the solid:
the tetrahedron in the first octant enclosed by the coordinate planes and the plane x+y+z=1.

Homework Equations

xcenter = $$\frac{\int\int\int_G x dV}{V}$$

ycenter = $$\frac{\int\int\int_G y dV}{V}$$

zcenter = $$\frac{\int\int\int_G z dV}{V}$$

The Attempt at a Solution

I have shown my attempt for xcenter, as the same problem arises for each one.

$$\frac{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} x dzdydx}{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV}$$

but $$\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV$$ is equal to zero,
so the above expression is undefined.

According to my text, the answer should be (1/4,1/4,1/4). Could someone point out what I did wrong? (Perhaps my bounds of integration?)

 

[tiny-tim]

$$\frac{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} x dzdydx}{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV}$$

but $$\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV$$ is equal to zero,
so the above expression is undefined.

Hi Knissp!

Try $$\int_{x=0}^1\int_{y=0}^ {1-x}\int_{z=0}^{1-y-x}$$

 

[Knissp]

Oh, I see now! Thank you!