Mass of a tetrahedron with density ρ = 24xyz

This problem asks to find the mass of a body with variable density using a triple integral.

Problem

Find the mass of the body bounded by the planes:

x = 0, y = 0, z = 0, x + y + z = 1

ρ (x, y, z) = 24 x y z

Solution

Step 1

The body V is a tetrahedron in the first octant with vertices O(0;0;0), A(1;0;0), B(0;1;0), C(0;0;1). It is bounded by the coordinate planes x = 0, y = 0, z = 0 and the plane x + y + z = 1.

Why is the body exactly a tetrahedron?

The plane x + y + z = 1 intersects the coordinate axes at (1;0;0), (0;1;0) and (0;0;1). Together with the origin (0;0;0), these four points form the vertices of a tetrahedron. All coordinates are non-negative, so the body lies in the first octant.

Step 2

Set the limits of integration. First determine the range in x, then in y (depending on x), then in z (depending on x and y):

0 \leq x \leq 1, 0 \leq y \leq 1 - x, 0 \leq z \leq 1 - x - y

Step 3

The mass of a body with variable density is computed via a triple integral:

m = \underset{V}{∭} ρ d V = \int_{0}^{1} d x \int_{0}^{1 - x} d y \int_{0}^{1 - x - y} 24 x y z d z

Step 4

Evaluate the inner integral over z (the density 24xy is moved outside as a constant with respect to z):

\int_{0}^{1 - x - y} 24 x y z d z = 24 x y \cdot \frac{{(1 - x - y)}^{2}}{2} = 12 x y {(1 - x - y)}^{2}

Step 5

Evaluate the middle integral over y. It is convenient to substitute u = 1 − x − y:

How is the substitution u = 1 − x − y performed?

With u = 1 − x − y we have y = 1 − x − u and dy = −du. The limits: for y = 0, u = 1 − x; for y = 1 − x, u = 0. The swap of limits together with the minus sign gives an integral from 0 to 1 − x in u. This substitution simplifies the power: u² remains instead of (1 − x − y)².

\int_{0}^{1 - x} 12 x y {(1 - x - y)}^{2} d y = x {(1 - x)}^{4}

Step 6

After substitution we obtain the outer integral over x:

m = \int_{0}^{1} x {(1 - x)}^{4} d x

Step 7

Evaluate this integral using the substitution t = 1 − x:

Why is the substitution t = 1 − x convenient?

The substitution t = 1 − x transforms the product x(1 − x)⁴ into (1 − t)t⁴ — a polynomial that integrates term by term. The limits swap (from 1 to 0), compensated by the minus sign from dt.

m = \int_{0}^{1} (1 - t) t^{4} d t = \int_{0}^{1} (t^{4} - t^{5}) d t = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}

Answer

m = 1/30.

Frequently asked questions

How do I find the mass of a body using a triple integral?

The mass is computed by m = ∭_V ρ(x, y, z) dV, where V is the volume of the body and ρ is the density. The triple integral is reduced to an iterated one by setting the limits along the boundaries: first over z (from the lower to the upper boundary), then over y (depending on x), then over x.

Why do the limits in y and z depend on other variables?

The body is bounded by the inclined plane x + y + z = 1. For fixed x, the value of y is bounded below by zero and above by 1 − x. Similarly, z is bounded by z = 0 and z = 1 − x − y. This is typical for regions with inclined boundaries.

Could another order of integration have been chosen?

Yes, one could integrate in a different order, for example over x, then z, then y. The result would be the same, but the computations might differ in complexity. In our case the order z → y → x naturally matches the boundaries of the body.

How do I verify the correctness of the answer?

One way is to estimate the average order of magnitude. The volume of the tetrahedron is V = 1/6. The average density 24xyz over the tetrahedron is about 24 · (1/4)³ = 24/64 = 3/8. Then the mass ≈ V · ρ̄ = (1/6)·(3/8) = 1/16. Our answer 1/30 is smaller, which is consistent: the density is maximal near the point (1;1;1), but in the tetrahedron this point is on the boundary, where the density does not accumulate.