Chapter 13: Q. 70 (page 1057)

Let a, b, and c be positive real numbers, and letR = {(x, y,z) | −a ≤ x ≤ a, −b ≤ y ≤ b, and −c ≤ z ≤ c}.
Prove that $∭_{R} α (x) β (y) γ (z) d V = 0$ if any of α, β, and γ is an odd function.

Short Answer

Expert verified

The given statement is proved.

Step by step solution

Step 1. Given Information.

It is given that $R = {(x, y, z) ∣ - a \leq x \leq a, - b \leq y \leq b, and - c \leq z \leq c} .$

Step 2. Prove.

To prove the given statement, we will use Fubini's theorem:

$∭_{R} α (x) β (y) γ (z) d v = \int_{- a}^{a} \int_{- b}^{b} [\int_{- c}^{c} α (x) β (y) γ (z) d z] d y d x ∭_{R} α (x) β (y) γ (z) d v = (\int_{- a}^{a} α (x) d x) (\int_{- b}^{b} β (y) d y) (\int_{- c}^{c} γ (z) d z)$

Now, as we know if f(x)is an odd function then $\int_{- a}^{a} f (x) = 0 .$

So, if any α, β, and γ is an odd function then one of the integral becomes zero.

So,

$(\int_{- a}^{a} α (x) d x) = 0$

$∭_{R} α (x) β (y) γ (z) d v = (0) (\int_{- b}^{b} β (y) d y) (\int_{- c}^{c} γ (z) d z) ∭_{R} α (x) β (y) γ (z) d v = 0$

Hence proved.

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Let a, b, and c be positive real numbers, and letR = {(x, y,z) | −a ≤ x ≤ a, −b ≤ y ≤ b, and −c ≤ z ≤ c}.
Prove that $∭_{R} α (x) β (y) γ (z) d V = 0$ if any of α, β, and γ is an odd function.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Prove.

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Let a, b, and c be positive real numbers, and letR = {(x, y,z) | −a ≤ x ≤ a, −b ≤ y ≤ b, and −c ≤ z ≤ c}.Prove that∭Rα(x)β(y)γ(z)dV=0 if any of α, β, and γ is an odd function.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Prove.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

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Study anywhere. Anytime. Across all devices.

Let a, b, and c be positive real numbers, and letR = {(x, y,z) | −a ≤ x ≤ a, −b ≤ y ≤ b, and −c ≤ z ≤ c}.
Prove that $∭_{R} α (x) β (y) γ (z) d V = 0$ if any of α, β, and γ is an odd function.