Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

Question

asked Nov 8, 2021 164k views

1 Answer

Radu Grama · Answer 1 · 2021-11-14T15:12:17+0000

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = (2)/(15)}$

Explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^(1-z)_0 \int ^(1-z)_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^(1-z)_0 ( ((1-z)^3)/(3)+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( ((1-z)^3)/(3) \ y + \frac {(1-z)y^3)}{3}] ^(1-x)_(0)$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( ((1-z)^4)/(3)+ ((1-z)^4)/(3)) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =(2)/(3) \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- (2)/(15)(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = (2)/(15)}$