Find the volume of the region by triple integral

Question

What is triple integral? How can I sketch region $D$ as well as evaluate it's volume? I get stuck.

Answer 1

By considering the $x$, $y$, and $z$ intercepts, we see that the region is the pyramid determined by the points $(1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1)$. By slicing the pyramid at a fixed $z$ and solving for $x$ and $y$ in terms of $z$, we obtain: $$ V = \int_0^1 \int_0^{1-z} \int_0^{1 - z} dx \, dy \, dz = \int_0^1 (1 - z)^2 \, dz = \left[ z - z^2 + \frac{z^3}{3} \right]_0^1 = \frac{1}{3} $$ which agrees with the usual formula for finding the volume of a pyramid: $$ V = \frac{1}{3}(\textsf{area of base})(\textsf{height}) = \frac{1}{3}(1^2)(1) = \frac{1}{3} $$