Chapter 13: Q. 34 (page 1004)

Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32
$\int \int_{R} x^{2} y d A W h e r e R = {(x, y) | - 1 \leq x \leq 0, 0 \leq y \leq 2}$

Short Answer

Expert verified

The value of integral using iteration method is $\int \int_{R} x^{2} y d A W h e r e R = {(x, y) | - 1 \leq x \leq 0, 0 \leq y \leq 2}$ is $\frac{2}{3} u n i t s$

Step by step solution

Given information

We are given an integral $\int \int_{R} x^{2} y d A W h e r e R = {(x, y) | - 1 \leq x \leq 0, 0 \leq y \leq 2}$

Evaluate the integral

We get,

$\int_{0}^{2} \int_{- 1}^{0} x^{2} y d x d y \int_{0}^{2} {[\frac{x^{3} y}{3}]}^{0}_{- 1} d y = \int_{0}^{2} \frac{y}{3} d y = \frac{1}{6} {[y^{2}]}^{2}_{0} = \frac{2}{3}$

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Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32
$\int \int_{R} x^{2} y d A W h e r e R = {(x, y) | - 1 \leq x \leq 0, 0 \leq y \leq 2}$

Short Answer

Step by step solution

Given information

Evaluate the integral

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Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32∫∫Rx2ydAWhereR={(x,y)|-1≤x≤0,0≤y≤2}

Short Answer

Step by step solution

Given information

Evaluate the integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Decision Maths

Probability and Statistics

Calculus

Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32
$\int \int_{R} x^{2} y d A W h e r e R = {(x, y) | - 1 \leq x \leq 0, 0 \leq y \leq 2}$