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Q. 73

Expert-verified
Found in: Page 1057

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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

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

See the 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.

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