Chapter 13: Q. 72 (page 1057)

Let $f (x)$ be an integrable function on the rectangular solid $R = \{(x, y, z) | a_{1} ⩽ x ⩽ a_{2}, b_{1} ⩽ y ⩽ b_{2}, c_{1} ⩽ z ⩽ c_{2}\}$ , and let $κ \in ℝ .$ Use the definition of the triple integral to prove that:
$∭_{R} κ f (x, y, z) d V = κ ∭_{R} f (x, y, z) d V .$

Short Answer

Expert verified

$∭_{R} f (x, y, z) d V = \lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V . R e p l a c e κ f (x, y, z) w i t h f (x, y, z) w e g e t, ∭_{R} κ f (x, y, z) d V = κ \lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V = κ (∭_{R} f (x, y, z) d V) T h e r e f o r e, ∭_{R} κ f (x, y, z) d V = κ ∭_{R} f (x, y, z) d V .$

Step by step solution

Step 1. Given Information.

Given: $R = \{(x, y, z) | a_{1} ⩽ x ⩽ a_{2}, b_{1} ⩽ y ⩽ b_{2}, c_{1} ⩽ z ⩽ c_{2}\}$ and $κ \in ℝ$ .

Step 2. Proof.

$A s w e k n o w f r o m t h e d e f i n i t i o n o f t r i p l e i n t e g r a l s : ∭_{R} f (x, y, z) d V = \lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V . R e p l a c e κ f (x, y, z) w i t h f (x, y, z) i n a b o v e, ∭_{R} κ f (x, y, z) d V = \lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} κ f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V A s w e k n o w t h e c o n s \tan t s g e t o u t f r o m s u m m a t i o n s a n d l i m i t s w e g e t, ∭_{R} κ f (x, y, z) d V = κ \lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V = κ (\lim_{∆ \to 0} \sum_{i = 1}^{l} \sum_{j = 1}^{m} \sum_{k = 1}^{n} f (x_{i}^{*}, y_{j}^{*}, z_{k}^{*}) d V) = κ (∭_{R} f (x, y, z) d V) T h e r e f o r e, ∭_{R} κ f (x, y, z) d V = κ ∭_{R} f (x, y, z) d V .$

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Let $f (x)$ be an integrable function on the rectangular solid $R = \{(x, y, z) | a_{1} ⩽ x ⩽ a_{2}, b_{1} ⩽ y ⩽ b_{2}, c_{1} ⩽ z ⩽ c_{2}\}$ , and let $κ \in ℝ .$ Use the definition of the triple integral to prove that:
$∭_{R} κ f (x, y, z) d V = κ ∭_{R} f (x, y, z) d V .$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

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Let f(x)be an integrable function on the rectangular solid R=x,y,z|a1⩽x⩽a2,b1⩽y⩽b2,c1⩽z⩽c2, and let κ∈ℝ.Use the definition of the triple integral to prove that:∭Rκf(x,y,z)dV=κ∭Rf(x,y,z)dV.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Mechanics Maths

Discrete Mathematics

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Let $f (x)$ be an integrable function on the rectangular solid $R = \{(x, y, z) | a_{1} ⩽ x ⩽ a_{2}, b_{1} ⩽ y ⩽ b_{2}, c_{1} ⩽ z ⩽ c_{2}\}$ , and let $κ \in ℝ .$ Use the definition of the triple integral to prove that:
$∭_{R} κ f (x, y, z) d V = κ ∭_{R} f (x, y, z) d V .$