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

#3x+2y+z=6#

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#

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 #