Calculate the volume integral of the function $T=z$²over the tetrahedron with comers at (0,0,0), (1,0,0), (0,1,0), and (0,0,1).

$1-y-z$The volume integral for a function $f(x,y,z)$over a volume V can be calculated as $\int \int \int$$f(x,y,z)dxdydz$. The given points are joined to make the following figure.

The figure obtained is of a tetrahedron, over which the volume integral has to be evaluated

From the figure, it can be inferred that plane formed by tetrahedron is $x+y+z=1$, as the points intercepts at 1 on each axis. Also, the x varies from 0 to $1-y-z$, y varies from 0 to $1-z$and z varies from 0 to 1.

The integral for a function can be calculated as,

$\int \int \int f(x,y)dxdydz$

Substitute $z$² for $f(x,y,z)$into the equation.

$I={\int }_0^1{\int }_{y=0}^{1-z}{\int }_{x=0}^{1-y-z}z$² $dxdydz$

$={\int }_0^1{\int }_{y=0}^{1-z}(1-y-z)z$²$dydz$

$={\int }_0^1(y-y$²$/2-yz)$^1-z₀ $z$² $dz$

$={\int }_0^1(1-z)-(1-z)$²$/2-z(1-z))z$² $dx$

Solve further as

$I={\int }_0^1(-3z+1/2+3z$²$/2)z$²

$={\int }_0^1\underset{2}{\underbrace{z}}$²$+\underset{2}{\underbrace{3z}}$⁴$-3z$³ $dz$

role="math" localid="1650071081839" $=(\underset{6}{\underbrace{z}}$³$+\underset{10}{\underbrace{3z}}$⁵ $-3z$⁴$/4)$¹₀

role="math" localid="1650071517054" $\frac{=-17}{60}$

Thus, the volume integral over the surface is .$=-17/60$