There are two ways do evaluate this volume.
The first is to use the high school formula for the volume of a pyramid. This is V = (1/3) base_area x height.
The base area (in the x y plane) is 1/2. The height = 3, so V = (1/3) (1/2) 3 = 1/2.
The calculus method is to notice that we have a stack of isosceles right triangles all normal to the z axis.
The side, s, of these triangles is s(z) = (3-z)/3. Since the area of an isosceles triangle of side s is
(1/2) s^{2 },^{
}
we must integrate (1/2) [ (3-z)/3 ]^{2} dz from 0 to 3. This integral is elementary because the integrand is
a polynomial form. It evaluates to 1/2 as expected.