Question

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A tetrahedron is bounded by the coordinate planes and the plane \(x / a+y / a+z / a=1 .\) What are the coordinates of the center of mass?

Short answer

Answer: The center of mass of the tetrahedron is at the coordinates $\left(\frac{1}{4}a, \frac{1}{4}b, \frac{1}{4}c\right)$.

Step-by-step solution

Express the tetrahedron as a solid

The given tetrahedron is bounded by the plane \(x / a + y / b + z / c = 1\). We can rewrite this as \(z = c(1 - x / a - y / b)\). Now, we can express the tetrahedron in Cartesian coordinates as the following region: $$ D = \{(x,y,z) \in \mathbb{R}^3 | 0 \leq x \leq a, \ 0 \leq y \leq b(1 - x / a), \ 0 \leq z \leq c(1 - x / a - y / b)\}. $$

Find the volume of the tetrahedron

To find the volume of the tetrahedron, we will compute the triple integral of the function 1 over the region D: $$ V = \iiint_{D} 1 dx dy dz = \int_{0}^{a} \int_{0}^{b(1-x/a)} \int_{0}^{c(1-x/a-y/b)} 1 dz dy dx. $$ Integrating with respect to z, we get: $$ V = \int_{0}^{a} \int_{0}^{b(1-x / a)} c(1 - x / a - y / b) dy dx. $$ Continuing with integration, first with respect to y and then x, we get: $$ V = abc \int_{0}^{a} (1 - x / a)^2 \frac{1}{2} dx = \frac{1}{6} abc \int_{0}^{a} (1 - x / a)^2 dx. $$ Integrating the last integral, we finally get the volume: $$ V = \frac{1}{6} abc \left. \left( \frac{x}{a} - \frac{x^2}{2a^2} \right) \right|_{0}^{a} = \frac{1}{6} abc. $$

Find the coordinates of the center of mass

The coordinates of the center of mass (X, Y, Z) are given by the following equations: $$ X = \frac{1}{V} \iiint_{D} x dx dy dz, \quad Y = \frac{1}{V} \iiint_{D} y dx dy dz, \quad Z = \frac{1}{V} \iiint_{D} z dx dy dz . $$ We will now compute these integrals one by one. To find X, we compute the triple integral: $$ \iiint_{D} x dx dy dz = \int_{0}^{a} x \int_{0}^{b(1 - x / a)} \int_{0}^{c(1 - x / a - y / b)}1dz dy dx. $$ Integrating with respect to z, y, and x, we get: $$ X = \frac{1}{V} \cdot \frac{1}{24} abc^2 = \frac{1}{4} a. $$ Next, we compute the triple integral for Y: $$ \iiint_{D} y dx dy dz = \int_{0}^{a} \int_{0}^{b(1 - x / a)} y \int_{0}^{c(1 - x / a - y / b)}1 dz dy dx. $$ Integrating with respect to z, y, and x, we get: $$ Y = \frac{1}{V} \cdot \frac{1}{24} ab^2 c = \frac{1}{4} b. $$ Lastly, we compute the triple integral for Z: $$ \iiint_{D} z dx dy dz = \int_{0}^{a} \int_{0}^{b(1 - x / a)} \int_{0}^{c(1 - x / a - y / b)} z dz dy dx. $$ Integrating with respect to z, y, and x, we get: $$ Z = \frac{1}{V} \cdot \frac{1}{24} a b c^2 = \frac{1}{4} c. $$

Write the coordinates of the center of mass

Now that we have the coordinates (X, Y, Z), we can write the center of mass of the tetrahedron as: $$ (X, Y, Z) = \left(\frac{1}{4}a, \frac{1}{4}b, \frac{1}{4}c\right). $$