A function is a pdf on the three-dimensional region Q if for all in Q and . Find k such that is a pdf on the region bounded by z = 0, z = 3, x = 0, x = 2, y - z = 0 , and y - z = 3.

Question

A function  is a pdf on the three-dimensional region Q if  for all  in Q and  . Find k such that  is a pdf on the region bounded by z = 0, z = 3, x = 0, x = 2, y - z = 0 , and y - z = 3.

Quizplus · Accepted Answer

To find the value of k, we need to use the property that the integral of the pdf over the entire region must equal 1. Mathematically, this is written as:

∫∫∫Q kxyz dV = 1

where Q is the region bounded by z = 0, z = 3, x = 0, x = 2, y - z = 0, and y - z = 3.

The limits of integration for x, y, and z can be obtained from the region boundaries:

0 ≤ x ≤ 2
0 ≤ y - z ≤ 3   (or z ≤ y ≤ z+3)
0 ≤ z ≤ 3

Thus, the integral can be rewritten as:

∫0^3 ∫0^2 ∫z^(z+3) kxyz dy dx dz

Evaluating this integral gives:

∫0^3 ∫0^2 kx/2 [(z+3)^2 - z^2] dx dz
= k/2 ∫0^3 [(z+3)^2 - z^2] ∫0^2 x dx dz
= k/2 ∫0^3 [(z+3)^2 - z^2] dz
= k/2 [∫0^3 (z+3)^2 dz - ∫0^3 z^2 dz]
= k/2 [(1/3)(6^3) - (1/3)(3^3)]
= k/2 (54 - 9)
= 22.5k

Setting this equal to 1, we get:

22.5k = 1
k = 1/22.5
k = 0.0444...

Therefore, the answer is A (0.0444...).

A Function Is a Pdf on the Three-Dimensional Region