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

2 Answers
Aug 8, 2016

= 6 =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:

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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

Aug 8, 2016

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