How would you find the volume of the tetrahedron T bounded by the coordinate planes and the plane 3x+4y+z=10?

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 + 4y + z = 10# is #(0, 0, 10)#, Similarly, the other three vertices are #(10/3, 0, 0)#, #(0, 5/2, 0)# and the origin #(0, 0, 0)#.

The given tetrahedron, #T#, is a solid that lies above the triangle #R# in the #xy#-plane that has vertices #(0, 0)#, #(10/3, 0)# and #(0, 5/2)#.

The line joining #(0, 5/2)# and #(10/3, 0)# is given by:

# y-0 = ((5/2)/(-10/3))(x-10/3) #
# :. y = -3/4x+5/2 #

And so the region #R# is defined as:

# R = {(x, y) | 0 le x le 10/3, 0 le y le −3/4x +5/2 } #

And the volume, #V#, of the tetrahedron is the double
integral of the function #z=10 − 3x − 4y# over #R#.

# :. V = int int_R (10 − 3x − 4y) \ dA#
# " "= int_0^(10/3) int_(0)^(-3/4x+5/2) (10 − 3x − 4y) \ dy \ dx#
# " "= int_0^(10/3) [(10-3x)y-2y^2]_(y=0)^(y=-3/4x+5/2) \ dx #
# " "= int_0^(10/3) {(10-3x)((-3x)/4+5/2)-2((-3x)/4+5/2)^2} - {0} \ dx #
# " "= int_0^(10/3) (-30/4x+25+9/4x^2-15/2x)-2(9/16x^2-15/4x+25/4) \ dx #
# " "= int_0^(10/3) (-15/2x+25/2+9/8x^2) \ dx #
# " "= [-15/4x^2+25/2x+9/24x^3]_0^(10/3) #
# " "= {-15/4*100/9+25/2*10/3+3/8*1000/27} - {0} #
# " "= -125/3+125/3+125/9 #
# " "= 125/9 #