Q. 70 - Calculus

Question

Let a, b, and c be positive real numbers, and let R = {(x, y,z) | −a ≤ x ≤ a, −b ≤ y ≤ b, and −c ≤ z ≤ c}.

Prove that${\iiint }_{\mathfrak{R}}\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dV=0$ if any of α, β, and γ is an odd function.

Step-by-Step Solution

Answer

The given statement is proved.

Step 1. Given Information.

It is given that $\mathfrak{R}=\left\{\right(x,y,z)\mid -a\le x\le a,-b\le y\le b\text{, and }-c\le z\le c\}\text{.}$

Step 2. Prove.

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

$$\begin{gathered} {\iiint }_{\mathfrak{R}}\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dv={\int }_{-a}^a{\int }_{-b}^b\left[{\int }_{-c}^c\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dz\right]dydx \\ {\iiint }_{\mathfrak{R}}\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dv=\left({\int }_{-a}^a\alpha \left(x\right)dx\right)\left({\int }_{-b}^b\beta \left(y\right)dy\right)\left({\int }_{-c}^c\gamma \left(z\right)dz\right) \end{gathered}$$

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

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

So,

$\left({\int }_{-a}^a\alpha \left(x\right)dx\right)=0$

$$\begin{gathered} {\iiint }_{\mathfrak{R}}\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dv=\left(0\right)\left({\int }_{-b}^b\beta \left(y\right)dy\right)\left({\int }_{-c}^c\gamma \left(z\right)dz\right) \\ {\iiint }_{\mathfrak{R}}\alpha \left(x\right)\beta \left(y\right)\gamma \left(z\right)dv=0 \end{gathered}$$

Hence proved.