Chapter 5: Q16P (page 247)

$\iint {(9 + 2 y^{2})}^{- 1}$ over the quadrilateral with vertice $(1, 3), (3, 3), (2, 6), (6, 6)$

Short Answer

Expert verified

The required solution is $\frac{In 3}{6}$

Step by step solution

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 $\iint_{A} f (x, y) d x d y$ .

Step 2: Drawing the area bounded by the curve

The area is bounded by the quadrilateral with vertices $(1, 3), (3, 3), (2, 6), (6, 6)$ .

Calculation of the area under the curve

First, integrate over x:

$\iint_{A} \frac{d y d x}{9 + 2 y^{2}} = \int_{y = 3}^{6} [\int_{x = y / 3}^{y} d x] \frac{d y}{9 + 2 y^{2}} = \int_{3}^{6} \frac{2}{3} y \frac{d y}{2 y^{2}}$

Further calculation of the area of the bounded region

Now, integrate over y:

$\frac{2}{3} \int_{3}^{6} \frac{y d y}{9 + 2 y^{2}} = \frac{1}{6} \int_{3}^{6} \frac{(d 9 + 2 y^{2})}{9 + 2 y^{2}} = \frac{1}{6} I n {(9 + 2 y^{2})}_{3}^{6} = \frac{I n 3}{6}$

Therefore, the value is $\frac{I n 3}{6}$

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$\iint {(9 + 2 y^{2})}^{- 1}$ over the quadrilateral with vertice $(1, 3), (3, 3), (2, 6), (6, 6)$

Short Answer

Step by step solution

Definition of double integral

Step 2: Drawing the area bounded by the curve

Calculation of the area under the curve

Further calculation of the area of the bounded region

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∬(9+2y2)-1 over the quadrilateral with vertice(1,3),(3,3),(2,6),(6,6)

Short Answer

Step by step solution

Definition of double integral

Step 2: Drawing the area bounded by the curve

Calculation of the area under the curve

Further calculation of the area of the bounded region

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Fields in Physics

Famous Physicists

Geometrical and Physical Optics

Space Physics

Electric Charge Field and Potential

Study anywhere. Anytime. Across all devices.

$\iint {(9 + 2 y^{2})}^{- 1}$ over the quadrilateral with vertice $(1, 3), (3, 3), (2, 6), (6, 6)$