91Ó°ÊÓ

The volume integral for a function $\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$ over a volume V can be calculated as $\underset{\mathbf{V}}{\mathbf{∭}}\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathit{d}\mathit{x}\mathit{d}\mathit{y}\mathit{d}\mathit{z}$.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 $f\left(x,y,z\right)$ can be calculated as,

$∭f\left(x,y,z\right) dx dy dz$

Substitute $z^2$for $f\left(x,y,z\right)$ into the equation.

$\begin{array}{rcl}I& =& \underset{0}{\overset{1}{∫}}\underset{y=0}{\overset{1-z}{∫}}\underset{x=0}{\overset{1-y-z}{∫}} z^{2 }dx dy dz\\ & =& \underset{0}{\overset{1}{∫}}\underset{y=0}{\overset{1-z}{∫}} \left(1-y-z\right)z^2  dy dz\\ & =& \underset{0}{\overset{1}{∫}}{\left(y-\frac{y^2}{2}-yz\right)}_0^{1-z}{z}^{2}dz\\ & =& \underset{0}{\overset{1}{∫}} \left(\left(1-z\right)-\frac{{\left(1-z\right)}^2}{2}-z\left(1-z\right)\right)z^2  dz\\ & & \end{array}$

Solve further as

$\begin{array}{rcl}I& =& \underset{0}{\overset{1}{∫}} \left(-3z+\frac{1}{2}+\frac{3z^2}{2}\right)z^2  dz\\ & =& \underset{0}{\overset{1}{∫}} \frac{z^2}{2}+\frac{3z^4}{2}-3z^3  dz\\ & =& {\left(\frac{z^3}{6}+\frac{3z^5}{10}-\frac{3z^4}{4}\right)}_0^1\\ & =& -\frac{17}{60}\\ & & \end{array}$

Thus, the volume integral over the surface T is $-\frac{17}{60}$.