Chapter 5: Q17P (page 247)

$\iint (\frac{x}{y})$ over the triangle with vertices $(0, 0), (1, 1), (1, 2)$

Short Answer

Expert verified

The required solution is $\frac{I n 2}{2}$ .

Step by step solution

Step 1: Definition of double integral

The double integral of $f (x, y)$ over the areain the $(x, y)$ plane as the limit of this sum, and we write it as $\iint_{A} f (x, y) d x d y$ .

Drawing the area bounded by the curve

The area is bounded by the triangle with vertices $(0, 0), (1, 1), (1, 2)$ ..

Calculation of the area under the curve

Integrating over the variable y

$\iint_{A} \frac{x}{y} d x d y = \int_{x = 0}^{1} [\int_{x}^{2 x} \frac{d y}{y}] x d x = \int_{0}^{1} (I {n y}_{x}^{2 x}) x d x = \int_{0}^{1} (I n (2 x) - I n x) x d x$

Further calculation

Integrating over the variable x

$\int_{0}^{1} (I n (2 x) - I n x) x d x = \int_{0}^{1} (I n 2 + I n x - I n x) x d x = {\frac{I n 2}{2} x^{2}}_{0}^{1} = \frac{I n 2}{2}$

Therefore, the value is $\frac{I n 2}{2}$

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$\iint (\frac{x}{y})$ over the triangle with vertices $(0, 0), (1, 1), (1, 2)$

Short Answer

Step by step solution

Step 1: Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve

Further calculation

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∬xyover the triangle with vertices (0,0),(1,1),(1,2)

Short Answer

Step by step solution

Step 1: Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve

Further calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Medical Physics

Classical Mechanics

Atoms and Radioactivity

Conservation of Energy and Momentum

Electromagnetism

Study anywhere. Anytime. Across all devices.

$\iint (\frac{x}{y})$ over the triangle with vertices $(0, 0), (1, 1), (1, 2)$