Q. 72 - Calculus

Question

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

${\iiint }_{\mathfrak{R}}\kappa f(x, y, z)dV = \kappa {\iiint }_{\mathfrak{R}}f(x, y, z)dV.$

Step-by-Step Solution

Answer

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

Step 1. Given Information.

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

Step 2. Proof.

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