Find the volume of the solid bounded by the planes x = 0, y = 0, and the plane z = 1 - x - y in the first octant.

Steps to Solve

1. Set Up the Integral To find the volume, we can use a triple integral. The volume $V$ can be expressed as: $$ V = \int_{0}^{1} \int_{0}^{1-x} (1 - x - y) , dy , dx $$ Here, the limits for $x$ are from 0 to 1, and for each $x$, $y$ ranges from 0 to $1 - x$.

2. Integrate with Respect to y First, we need to integrate the inner integral with respect to $y$: $$ \int_{0}^{1-x} (1 - x - y) , dy $$ Calculating this gives: $$ = \left[ (1 - x)y - \frac{y^2}{2} \right]_{0}^{1-x} $$ Substituting the limits in yields: $$ = (1 - x)(1 - x) - \frac{(1 - x)^2}{2} $$

3. Simplify the Expression Combine and simplify the result: $$ = (1 - x)^2 - \frac{(1 - x)^2}{2} $$ Factor out $(1 - x)^2$: $$ = (1 - x)^2 \left(1 - \frac{1}{2}\right) = \frac{(1 - x)^2}{2} $$

4. Integrate with Respect to x Now, we integrate this result with respect to $x$ from 0 to 1: $$ V = \int_{0}^{1} \frac{(1 - x)^2}{2} , dx $$ Calculating this gives: $$ = \frac{1}{2} \int_{0}^{1} (1 - 2x + x^2) , dx $$ Integrating each term: $$ = \frac{1}{2} \left[ x - x^2 + \frac{x^3}{3} \right]_{0}^{1} $$

5. Evaluate the Limits Substituting the limits in: $$ = \frac{1}{2} \left(1 - 1 + \frac{1}{3}\right) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} $$