Chapter 13: Q. 41 (page 1079)

Evaluate the double integrals in Exercises 39–48. Use suitable transformations as necessary.
$\int \int_{R} \frac{2 y - x}{3 x + y + 1} d A$ , where R is the parallelogram with vertices (0, 0), (2, 1), (1, 4), and (−1, 3) .

Short Answer

Expert verified

$\int \int_{R} \frac{2 y - x}{3 x + y + 1} d A = \frac{21}{2} \ln (2)$

Step by step solution

Draw the region

Plot the given points to form the region and name the vertices.

Consider the new set of variables defined as,

$u = x - 2 y v = 3 x + y$

From the above equations,

$\frac{u + 2 v}{7} = x \frac{v - 3 u}{7} = y$

Plot these limits on u v plane,

Find the integration

Use the antiderivative to evaluate the integrals,

$\int \int_{R} \frac{2 y - x}{3 x + y + 1} d A = - \frac{3}{7} \ln (2) \int_{- 7}^{0} u d u = - \frac{3}{7} \ln (2) {[\frac{u^{2}}{2}]}_{- 7}^{0} = - \frac{3}{7} \ln (2) [- \frac{49}{2}] = \frac{21}{2} \ln (2)$

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Evaluate the double integrals in Exercises 39–48. Use suitable transformations as necessary.
$\int \int_{R} \frac{2 y - x}{3 x + y + 1} d A$ , where R is the parallelogram with vertices (0, 0), (2, 1), (1, 4), and (−1, 3) .

Short Answer

Step by step solution

Draw the region

Find the integration

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Evaluate the double integrals in Exercises 39–48. Use suitable transformations as necessary.∫∫R2y-x3x+y+1dA, where R is the parallelogram with vertices (0, 0), (2, 1), (1, 4), and (−1, 3) .

Short Answer

Step by step solution

Draw the region

Find the integration

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Decision Maths

Geometry

Pure Maths

Statistics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Evaluate the double integrals in Exercises 39–48. Use suitable transformations as necessary.
$\int \int_{R} \frac{2 y - x}{3 x + y + 1} d A$ , where R is the parallelogram with vertices (0, 0), (2, 1), (1, 4), and (−1, 3) .