Q. 73 - Calculus

Question

Let f(x, y, z) and g(x, y, z) be integrable functions on the rectangular solid $\mathfrak{R}=\left\{\left(x, y, z\right) | a_1⩽x⩽a_2, b_1⩽y⩽b_2, c_1⩽z⩽c_2\right\}$. . Use the definition of the triple integral to prove that :

${\iiint }_{\mathfrak{R}}\left(f(x, y, z)+g(x, y, z)\right)dV = {\iiint }_{\mathfrak{R}}f(x, y, z)dV+{\iiint }_{\mathfrak{R}}g(x, y, z)dV.$

Step-by-Step Solution

Answer

$$\begin{gathered} {\iiint }_{\mathfrak{R}}f(x, y, z)dV = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}f(x_i^{*}, y_j^{*}, z_k^{*})dV. \\ Replace f(x, y, z)+g(x, y, z) with f(x, y, z) in above, \\ = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}f(x_i^{*}, y_j^{*}, z_k^{*})dV + \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}g(x_i^{*}, y_j^{*}, z_k^{*})dV \\ Therefore, \\ {\iiint }_{\mathfrak{R}}\left(f(x, y, z)+g(x, y, z)\right)dV = {\iiint }_{\mathfrak{R}}f(x, y, z)dV+{\iiint }_{\mathfrak{R}}g(x, y, z)dV. \\ Hence, proved. \end{gathered}$$

Step 1. Given Information.

Given: $\mathfrak{R}=\left\{\left(x, y, z\right) | a_1⩽x⩽a_2, b_1⩽y⩽b_2, c_1⩽z⩽c_2\right\}$

Step 2. Proof.

$$\begin{gathered} As we know from the definition of triple integrals: \\ {\iiint }_{\mathfrak{R}}f(x, y, z)dV = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}f(x_i^{*}, y_j^{*}, z_k^{*})dV. \\ Replace f(x, y, z)+g(x, y, z) with f(x, y, z) in above, \\ {\iiint }_{\mathfrak{R}}\left(f(x, y, z)+g(x, y, z)\right)dV = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^n\left(f+g\right)(x_i^{*}, y_j^{*}, z_k^{*})dV \\ = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^n\left(f(x_i^{*}, y_j^{*}, z_k^{*})+g(x_i^{*}, y_j^{*}, z_k^{*})\right)dV \\ = \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}f(x_i^{*}, y_j^{*}, z_k^{*})dV + \underset{∆\to 0}{\lim }\sum _{i=1}^l\sum _{j=1}^m\sum _{k=1}^{n}g(x_i^{*}, y_j^{*}, z_k^{*})dV \\ ={\iiint }_{\mathfrak{R}}f(x, y, z)dV +{\iiint }_{\mathfrak{R}}g(x, y, z)dV = RHS \\ Therefore, \\ {\iiint }_{\mathfrak{R}}\left(f(x, y, z)+g(x, y, z)\right)dV = {\iiint }_{\mathfrak{R}}f(x, y, z)dV+{\iiint }_{\mathfrak{R}}g(x, y, z)dV. \\ Hence, proved. \end{gathered}$$