Problem Setup: Bounded Solid Volume

The problem asks for the volume of a solid region in the first octant ($x \ge 0, y \ge 0, z \ge 0$) bounded by the coordinate planes ($x=0$, $y=0$, $z=0$) and the plane $x+y+z=2$.

The plane $x+y+z=2$ intersects the $z$-axis at $z=2$, the $y$-axis at $y=2$, and the $x$-axis at $x=2$. The solid is a tetrahedron with vertices at (0,0,0), (2,0,0), (0,2,0), and (0,0,2).

We can find the volume using a double integral. The upper surface is given by $z = 2 - x - y$. The region of integration $R$ in the $xy$-plane is determined by the intersection of the solid with the plane $z=0$, which gives the triangle defined by $x \ge 0$, $y \ge 0$, and $x+y \le 2$.

Integral Formulation

The volume $V$ can be computed as:

$V = \iint_R z \, dA = \iint_R (2 - x - y) \, dA$

The region $R$ can be described by the limits:

• $0 \le x \le 2$
• $0 \le y \le 2 - x$

The integral becomes:

$V = \int_0^2 \int_0^{2-x} (2 - x - y) \, dy \, dx$

Volume Calculation

First, evaluate the inner integral with respect to $y$:

$\int_0^{2-x} (2 - x - y) \, dy = \left[ (2 - x)y - \frac{y^2}{2} \right]_0^{2-x}$

= $\left( (2 - x)(2 - x) - \frac{(2 - x)^2}{2} \right) - (0 - 0)$

= $(2 - x)^2 - \frac{1}{2}(2 - x)^2 = \frac{1}{2}(2 - x)^2$

Now, evaluate the outer integral with respect to $x$:

$V = \int_0^2 \frac{1}{2}(2 - x)^2 \, dx$

Let $u = 2 - x$. Then $du = -dx$. The limits change: when $x=0$, $u=2$; when $x=2$, $u=0$.

$V = \int_2^0 \frac{1}{2} u^2 (-du) = \int_0^2 \frac{1}{2} u^2 \, du$

$V = \frac{1}{2} \left[ \frac{u^3}{3} \right]_0^2 = \frac{1}{2} \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$

$V = \frac{1}{2} \left( \frac{8}{3} \right) = \frac{4}{3}$

The volume of the solid is $\frac{4}{3}$.