Question

How do you find the volume of the solid bounded by the coordinate planes and the plane `3x+2y+z=1`?

Answer 1

Volume = 1/36 (`"units"^3`)

Explanation:

The coordinate planes are given by `x = 0`, `y = 0` and `z = 0`. The volume is that of a tetrahedron whose vertices are the intersections of three of the four planes given. The intersection of `x = 0`, `y = 0` and `3x + 2y + z = 1` is `(0, 0, 1)`, Similarly, the other three vertices are `(1/3, 0, 0)`, `(0, 1/2, 0)` and the origin `(0, 0, 0)`.

The given tetrahedron is a solid that lies above the triangle `R` in the `xy`-plane that has vertices `(0, 0)`, `(1/3, 0)` and `(0, 1/2)`.

The line joining `(0, 1/2)` and `(1/3, 0)` is given by:

`y-0 = ((1/2)/(-1/3))(x-1/3)`
`:. y = -3/2x+1/2`

And so the region `R` is defined as:

`R = {(x, y) | 0 le x le 1/3, 0 le y le −3/2x +1/2 }`

And the volume, `V`, of the tetrahedron is the double
integral of the function `z=1 − 3x − 2y` over `R`.

`:. V = int int_R (1 − 3x − 2y) \ dA`
`" "= int_0^(1/3) int_(0)^(-3/2x+1/2) (1 − 3x − 2y) \ dy \ dx`
`" "= int_0^(1/3) [(1-3x)y-y^2]_(y=0)^(y=-3/2x+1/2) \ dx`
`" "= int_0^(1/3) {(1-3x)((-3x)/2+1/2)-((-3x)/2+1/2)^2} - {0} \ dx`
`" "= int_0^(1/3) (-3/2x+1/2+9/2x^2-3/2x)-(9/4x^2-3/2x+1/4) \ dx`
`" "= int_0^(1/3) (1/4-3/2x+9/4x^2) \ dx`
`" "= [1/4x-3/4x^2+9/12x^3]_0^(1/3)`
`" "= {1/12-1/12+1/36} - {0}`
`" "= 1/36`