Volume of Tetrahedron

Geometry Level pending

What is the volume of the tetrahedron bounded by the planes x + 2 y + z = 2 x+2y+z=2 , x = 2 y x=2y , x = 0 x=0 , and z = 0 z=0 ?

1 1 3 \frac{1}{3} 1 6 \frac{1}{6} 2 3 \frac{2}{3}

Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 14, 2020

In the x y − xy- plane ( z = 0 z=0 ) we have the triangular region bounded by x = 0 , y = x 2 , y = − x 2 + 1 x=0, y=\frac{x}{2}, y = -\frac{x}{2} +1 , which has vertices at ( 0 , 0 , 0 ) ; ( 2 , 0 , 0 ) ; ( 1 , 1 / 2 , 0 ) (0,0,0); (2,0,0); (1, 1/2, 0) . The area of this triangular region computes to:

A = 1 2 ⋅ ∣ 1 0 0 1 2 0 1 1 1 / 2 ∣ = 1 2 A = \frac{1}{2} \cdot \begin{vmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1/2 \end{vmatrix} = \frac{1}{2} .

The volume of the tetrahedron finally calculates according to V = 1 3 A h = 1 3 ⋅ 1 2 ⋅ 2 = 1 3 . V =\frac{1}{3}Ah = \frac{1}{3} \cdot \frac{1}{2} \cdot 2 = \boxed{\frac{1}{3}}.

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