How do you use a triple integral to find the volume of the given the tetrahedron enclosed by the coordinate planes 2x+y+z=3?

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Answer 1 · 2016-06-30T11:35:32

# = 9/4#

we can start by writing the plane as #z = 3 - 2x - y#

so by setting z = 0, we know that the place cuts across the x-y plane on 2x + y = 3

by setting (x,y) = 0, we know that the plane hits the z axis at z = 3

so we can draw it all in the first octant

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The triple Integral is

#int_{x = 0}^{3/2} int_{y=0}^{3-2x } int_{z=0}^{3 - 2x - y} dz dy dx#

#= int_{x = 0}^{3/2} int_{y=0}^{3-2x } 3 - 2x - y \ dy dx#

#= int_{x = 0}^{3/2} [3y - 2xy - y^2/2]_{y=0}^{3-2x } dx#

#= 1/2 int_{x = 0}^{3/2} (3-2x)^2 dx#

#= 1/2[ 1/3 (- 1/2)(3-2x)^3]_{x = 0}^{3/2} #

#= 1/12[(3-2x)^3]_{x = 3/2}^{0} #

# = 9/4#