Question

How do you find the volume of the solid bounded by the coordinate planes and the plane 3x + 2y + z = 6?

Answer 1

The answer is `=6 (unit)^2`

Explanation:

We have here a tetrahedron.

`3x+2y+z=6`

Let's find the vertices,

Let `y=0` and `z=0`,

we get `3x=6`, `=>`, `x=2` and

vertex `veca=〈2,0,0〉`

Let `x=0` and `z=0`

We get `2y=6`, `=>`, `y=3` and

vertex `vecb=〈0,3,0〉`

Let `x=0` and `y=0`

We get `z=6`

vertex `vecc=〈0,0,6〉`

And the volume is `V=1/6*∣veca.(vecbxxvecc)∣`

Where, `veca.(vecbxxvecc)` is the scalar triple product

`V=1/6*| (2,0,0), (0,3,0), (0,0,6) |`

`=1/6*2*3*6=6 (unit)^2`

Answer 2

Volume = 6 (`"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 = 6` is `(0, 0, 6)`, Similarly, the other three vertices are `(2, 0, 0)`, `(0, 3, 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)`, `(2, 0)` and `(0, 3)`.

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

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

And so the region `R` is defined as:

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

And the volume, `V`, of the tetrahedron is the double
integral of the function `z=6 − 3x − 2y` over `R`.

`:. V = int int_R (6 − 3x − 2y) \ dA`
`" "= int_0^2 int_0^(-3/2x+3) (6 − 3x − 2y) \ dy \ dx`
`" "= int_0^2 [(6-3x)y-y^2]_(y=0)^(y=-3/2x+3) \ dx`
`" "= int_0^2 {(6-3x)((-3x)/2+3)-((-3x)/2+3)^2} - {0} \ dx`
`" "= int_0^2 (-9x+18+9/2x^2-9x)-(9/4x^2-9x+9) \ dx`
`" "= int_0^2 (-9x+9+9/4x^2) \ dx`
`" "= [-9/2x^2+9x+9/12x^3]_0^2`
`" "= {-18+18+72/12} - {0}`
`" "= 6`