Find the volume of the tetrahedron bounded by the plane passing through the points (1,0,0), (0,2,0) and (0,0,3) and the coordinate planes.

Question

Find the volume of the tetrahedron bounded by the plane passing through the points (1,0,0), (0,2,0) and (0,0,3) and the coordinate planes.

Answer 1

Hi Ayaka,

Thanks for your question!

The first thing we need to determine is the equation of the plane that bounds the top surface of the tetrahedron. This plane will have the equation

z = A x + B y + C,

where A, B, and C are constants to be determined. We can obtain the value of these constants by requiring this plane to pass through the points (1,0,0), (0,2,0), and (0,0,3). This gives us a set of three equations for three unknowns:

0 = A + C,

0 = 2B + C,

3 = C.

Solving this system, we find A = -3, B = -3/2, and C = 3, so our plane of interest is

z = -3*x -3/2*y + 3.

To compute the volume of this tetrahedron requires us to compute the volume underneath this plane that lies in the first octant of 3D space (i.e., the octant with positive x, positive y, and positive z coordinates).

To compute the volume underneath this plane requires the double integral

Volume of Tetrahedron = ∫∫_R(-3*x -3/2*y + 3)dA,

where R is the region of the xy-plane over which our tetrahedron is defined.

A quick sketch of our tetrahedron shows that R is the finite region in the first quadrant of the xy-plane bounded by the x-axis, y-axis, and the line y = -2x + 2. (This line connects the points (1,0) and (0,2), which come from the points (1,0,0) and (0,2,0) on the tetrahedron.)

Given R, we can now set up our integral. Importantly, we must decide whether to use dA = dxdy or dA = dydx. Either is acceptable, but one is slightly simpler than the other. In this case, the preferred choice is dA = dydx, since one of the boundaries of our region R is expressed as y as a function of x (namely, y = -2x + 2). This gives us the integral

Volume of Tetrahedron = ∫∫_R(-3*x -3/2*y + 3)dydx.

Collapsing our region R in the xy-plane onto the x-axis gives the bounds of the outer integral with respect to x. We find that the lower bound is x = 0 and the upper bound is x = 1. To get the bounds of the inner integral with respect to y, we imagine fixing a value of x between 0 and 1 and assessing the vertical extent of our region R for that fixed x value. We see that the vertical extent of R for fixed x has lower bound y = 0 and upper bound y = -2x + 2. Hence, our integral with bounds becomes

Volume of Tetrahedron = ∫₀¹∫₀^-2x+2(-3*x -3/2*y + 3)dydx

= ∫₀¹[-3xy -3/4*y² + 3y]^-2x+2₀dx

= ∫₀¹(-3x(-2x+2)-3/4*(-2x+2)² + 3(-2x+2))dx

= [x³ - 3x² + 3x]¹₀

= 1.