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Math
Calculus
Chapter 13
Evaluating iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals.

Chapter 13: Evaluating iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals. (page 1082)

Q. 7 $\int_{0}^{1} \int_{- \sqrt{1 - y^{2}}}^{\sqrt{1 + y^{2}}} \frac{x}{y + 1} d x d y$

Short Answer

Expert verified

$\int_{0}^{1} \int_{- \sqrt{1 - y^{2}}}^{\sqrt{1 + y^{2}}} \frac{x}{y + 1} d x d y = 0$

Step by step solution

01

Draw the region

The region determined by the limits of the given integral is shown below,

02

Evaluate integral

Here a concept is used,

$\int_{- a}^{a} f (x) d x = 0, If f (x) is an odd function i . e . f (- x) = - f (x)$

$\int_{0}^{1} \int_{- \sqrt{1 - y^{2}}}^{\sqrt{1 + y^{2}}} \frac{x}{y + 1} d x d y = \int_{0}^{1} \frac{1}{y + 1} [\int_{- \sqrt{1 - y^{2}}}^{\sqrt{1 + y^{2}}} x d x] d y = \int_{0}^{1} \frac{1}{y + 1} [0] d y [f (x) = x, is an odd function] = 0$

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Most popular questions from this chapter

In Exercises 45–52, rewrite the indicated integral with the specified order of integration.

Exercise 41 with the order dy dx dz.

Evaluate the iterated integral :

$\int_{0}^{π / 2} \int_{0}^{\cos y} \int_{0}^{\sin y} (2 x + y) d z d x d y$

Use Definition $13.4$ to evaluate the double integrals in Exercises $29 - 32$ .

localid="1649936867482" $\iint_{x} x^{2} y d A$

where $R = {(x, y) ∣ 1 \leq x \leq 3 and 0 \leq y \leq 2}$

Describe the three-dimensional region expressed in each iterated integral:

$\int_{- 2}^{4} \int_{2}^{6} \int_{0}^{5} f (x, y, z) d y d x d z$

Evaluate each of the double integrals in Exercises $37 - 54$ as iterated integrals.

role="math" localid="1650327788023" $\int \int_{R} y \sin x d A,$

whererole="math" localid="1650327080219" $R = \{(x, y) | 0 \leq x \leq \frac{π}{2} a n d 0 \leq y \leq 1\}$

See all solutions

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