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Chapter 5: Q29P (page 248)

$\int_{y = 0}^{ττ} \int_{x = y}^{ττ} dxdy$

Short Answer

Expert verified

The required solution is 2.

Step by step solution

01

Definition of double integral

The double integral of f(x,y) over the area Ain the (x,y) plane as the limit of this sum, and we write it as $\int \int_{A} f (x, y) dxdy$

02

Drawing the area bounded by the curve

The area of integration or bounded region is shown below,.

03

Calculation of the area under the curve changing boundaries

Changing the boundaries of the given integration:

$x : 0 \to ττ, a and y : 0 \to x$

So, the integration

$l = \int_{0}^{π} \frac{\sin x}{x} dx \int_{0}^{x} dy = \int_{0}^{π} \frac{\sin x}{x} x = \int_{0}^{π} sinxdx = {- \cos x}_{0}^{π} = 2$

Therefore, the integration value is, 2.

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Most popular questions from this chapter

In Problems 17 to 30, for the curve $y = \sqrt{x}$ , between $x = 0$ and $x = 2$ ,

find:

The area under the curve.

$\int_{x = 1}^{2} \int_{y = x}^{2 x} \int_{z = 0}^{y - x} d z d y d x$

Question: $\int \int_{A} sinxdxdy$ where A is the area shown in Figure 2.8

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

(a) Revolve the curve $y = x^{1}$ , from $x = 1 to x = \infty$ , about the x axis to create a surface and a volume. Write integrals for the surface area and the volume. Find the volume, and show that the surface area is infinite. Hint: The surface area integral is not easy to evaluate, but you can easily show that it is greater than $\int_{1}^{\infty} x^{- 1} dx$ which you can evaluate.

(b) The following question is a challenge to your ability to fit together your mathematical calculations and physical facts: In (a) you found a finite volume and an infinite area. Suppose you fill the finite volume with a finite amount of paint and then pour off the excess leaving what sticks to the surface. Apparently, you have painted an infinite area with a finite amount of paint! What is wrong? (Compare Problem 15.31c of Chapter 1.)

See all solutions

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