Chapter 13: Q. 17 (page 1004)

Earlier in this section, we showed that we could use Fubini’s theorem to evaluate the integral $\int \int_{R} x^{2} y d A$ and we showed that $\int_{2}^{5} \int_{1}^{3} x^{2} y d x d y = 91$ Now evaluate the double integral by evaluating the iterated integral $\int_{1}^{3} \int_{2}^{5} x^{2} y d y d x$ .

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The solution is,

$\int_{1}^{3} \int_{2}^{5} x^{2} y d y d x = 91$ .

Step by step solution

Step 1. Given Information.

The iterated integral is $\int_{1}^{3} \int_{2}^{5} x^{2} y d y d x$ .

Step 2. Evaluating the integral.

We integrate with respect to $x$ .

$\int_{1}^{3} \int_{2}^{5} x^{2} y d y d x = \int_{1}^{3} (\int_{2}^{5} x^{2} y d y) d x = \int_{1}^{3} (\int_{2}^{5} x^{2} (y d y)) d x = \int_{1}^{3} {[x^{2} \frac{y^{1 + 1}}{1 + 1}]}_{y = 2}^{y = 5} d x = \int_{1}^{3} {[\frac{1}{2} x^{2} y^{2}]}_{y = 2}^{y = 5} d x = \int_{1}^{3} (\frac{1}{2} x^{2} 5^{2} - \frac{1}{2} x^{2} 2^{2}) d x = \int_{1}^{3} (\frac{25}{2} x^{2} - 2 x^{2}) d x = \int_{1}^{3} (\frac{21}{2} x^{2}) d x = \frac{21}{2} \int_{1}^{3} x^{2} d x = \frac{21}{2} {[\frac{x^{2 + 1}}{2 + 1}]}_{x = 1}^{x = 3} = \frac{21}{2} {[\frac{x^{3}}{3}]}_{x = 1}^{x = 3} = \frac{21}{2} [\frac{3^{3}}{3} - \frac{1^{3}}{3}] = \frac{21}{2} [\frac{27}{3} - \frac{1}{3}] = \frac{21}{2} \times \frac{26}{3} = 7 \times 13 = 91$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Evaluating the integral.

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Earlier in this section, we showed that we could use Fubini’s theorem to evaluate the integral ∫∫Rx2ydAand we showed that ∫25∫13x2ydxdy=91Now evaluate the double integral by evaluating the iterated integral∫13∫25x2ydydx.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Evaluating the integral.

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Most popular questions from this chapter

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