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 # 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#