Question

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)

Answer 1

`\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)}`