Question

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

Answer 1

`= 6` cubic units

Explanation:

the normal vector is `((2),(3),(1))` which points out in the direction of octant 1, so the volume in question is under the plane and in octant 1

we can re-write the plane as `z(x,y)= 6 - 2x - 3y`

for `z = 0` we have

• `z= 0, x = 0 implies y = 2`
• `z= 0, y = 0 implies x = 3`

and
- - `x= 0, y = 0 implies z = 6`

it's this:

the volume we need is

`int_A z(x,y) dA`

`= int_(x=0)^(3) int_(y=0)^(2 - 2/3 x) 6 - 2x - 3y \ dy \ dx`

`= int_(x=0)^(3) [ 6y - 2xy - 3/2y^2 ]_(y=0)^(2 - 2/3 x) \ dx`

`= int_(x=0)^(3) [ 6(2-2/3 x) - 2x(2-2/3 x) - 3/2(2-2/3 x)^2 ]_(y=0)^(2 - 2/3 x) \ dx`

`= int_(x=0)^(3) 12-4 x - 4x + 4/3 x^2 - 6 - 2/3 x^2 + 4x \ dx`

`= int_(x=0)^(3) 6- 4 x + 2/3 x^2 \ dx`

`=[ 6x- 2 x^2 + 2/9 x^3 ]_(x=0)^(3)`

`= 18- 18 + 54/9`

`= 6`

Answer 2

6

Explanation:

We are going to be performing a triple integral.
The cartesian coordinate system is the most applicable. The order of integration is not critical. We are going to go z first, y middle, x last.

`underline("Determining limits")`

On the plane `z = 6 - 2x - 3y` and on the coordinate plane `z = 0` hence

`z: 0 rarr 6 - 2x - 3y`

Along `z=0`, `y` goes from 0 to `3y = 6 - 2x` hence

`y: 0 rarr 2 - 2/3x`

Along `y=0, z=0` hence

`x: 0 rarr 3`

We are finding the volume so `f(x,y,z) = 1`. Integral becomes

`int_0^3int_0^(2-2/3x)int_0^(6-2x-3y)dzdydx`

`=int_0^3int_0^(2-2/3x)[z]_0^(6-2x-3y)dydx`

`=int_0^3int_0^(2-2/3x)(6-2x-3y)dydx`

`=int_0^3[6y-2xy - 3/2y^2]_0^(2-2/3x)dx`

`=int_0^3(6(2-2/3x) - 2x(2-2/3x) - 3/2(2-2/3x)^2) dx`

`=int_0^3 (12 - 4x - 4x + 4/3x^2 - 3/2(4 - 8/3x + 4/9x^2)) dx`

`=int_0^3 (12 - 8x + 4/3x^3 - 6 + 4x - 2/3x^2)dx`

`=int_0^3(6 - 4x + 2/3x^2)dx`

`= [6x - 2x^2 + 2/9x^3]_0^3`

`=6(3) - 2(3)^2 +2/9(3)^3`

`=6`