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 :

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

Short answer

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

Step-by-step solution

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.

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