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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∭Rα(x)β(y)γ(z)dV=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

01

Step 1. Given Information.

It is given thatR={(x,y,z)∣−a≤x≤a,−b≤y≤b, and−c≤z≤c}.

02

Step 2. Prove.

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

∭Rα(x)β(y)γ(z)dv=∫-aa∫-bb∫-ccα(x)β(y)γ(z)dzdydx∭Rα(x)β(y)γ(z)dv=∫-aaα(x)dx∫-bbβ(y)dy∫-ccγ(z)dz

Now, as we know if f(x)is an odd function then ∫-aaf(x)=0.

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

So,

∫-aaα(x)dx=0

∭Rα(x)β(y)γ(z)dv=0∫-bbβ(y)dy∫-ccγ(z)dz∭Rα(x)β(y)γ(z)dv=0

Hence proved.

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