Chapter 13: Q. 69 (page 1057)

Let $R = \{(x, y, z) | a_{1} \leq x \leq a_{2}, b_{1} \leq y \leq b_{2}, and c_{1} \leq z \leq c_{2}\} .$ If α(x), β( y), and γ (z) are integrable on the intervals $[a_{1}, a_{2}], [b_{1}, b_{2}], and [c_{1}, c_{2}],$ respectively, use Fubini’s theorem to prove that
$∭_{R} α (x) β (y) γ (z) d V = (\int_{a_{1}}^{a_{2}} α (x) d x) (\int_{b_{1}}^{b_{2}} β (y) d y) (\int_{c_{1}}^{c_{2}} γ (z) d z) .$

Short Answer

Expert verified

The given statement is proved.

Step by step solution

Step 1. Given Information.

It is given that $R = \{(x, y, z) | a_{1} \leq x \leq a_{2}, b_{1} \leq y \leq b_{2}, and c_{1} \leq z \leq c_{2}\} .$

Step 2. Prove.

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

$∭_{R} α (x) β (y) γ (z) d v = \int_{a_{1}}^{a_{2}} \int_{b_{1}}^{b_{2}} [\int_{c_{1}}^{c_{2}} α (x) β (y) γ (z) d z] d y d x ∭_{R} α (x) β (y) γ (z) d v = \int_{a_{1}}^{a_{2}} \int_{b_{1}}^{b_{2}} [α (x) β (y) \int_{c_{1}}^{c_{2}} γ (z) d z] d y d x$

Now,

role="math" localid="1650361075356" $= \int_{a_{1}}^{a_{2}} [\int_{b_{1}}^{b_{2}} α (x) β (y) (\int_{c_{1}}^{c_{2}} γ (z) d z) d y d x] = (\int_{a_{1}}^{a_{2}} α (x) d x) (\int_{b_{1}}^{b_{2}} β (y) d y) (\int_{c_{1}}^{c_{2}} γ (z) d z))$

Hence proved.

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Short Answer

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Step 1. Given Information.

Step 2. Prove.

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Let R=(x,y,z)|a1≤x≤a2,b1≤y≤b2,andc1≤z≤c2.If α(x), β( y), and γ (z) are integrable on the intervals [a1,a2],[b1,b2],and[c1,c2],respectively, use Fubini’s theorem to prove that∭Rα(x)β(y)γ(z)dV=∫a1a2α(x)dx∫b1b2β(y)dy∫c1c2γ(z)dz.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Prove.

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Most popular questions from this chapter

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Applied Mathematics

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