Chapter 5: Q28P (page 248)

$\int_{y = 0}^{1} \int_{x = 0}^{\sqrt{1 - y^{2}}} ydxdy$

Short Answer

Expert verified

The required solution is $\frac{1}{3}$

Step by step solution

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 $\int \int_{A} f (x, y) dxdy$

Drawing the area bounded by the curve

The area of integration or bounded region is shown below,

Calculation of the area under the curve in an integral way

At first, the integral is the way it is set up in the problem:

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

Let's assume,

$y = \sin u d y = \cos u d u$

Therefore,

$l = \int_{0}^{π / 2} \cos u d u \sin u \cos u = - \int_{0}^{π / 2} \cos^{2} u d (\cos u) = {- \frac{1}{3} \cos^{3} u}_{0}^{π / 2} = \frac{1}{3}$

Calculation of the area under the curve changing boundaries

The other way, changing the boundaries of the given integration:

$x : 0 \to 1, and y : 0 \to \sqrt{1 - x^{2}}$

So, the integration

$l = \int_{0}^{1} d x \int_{y^{2}}^{\sqrt{1 - x^{2}}} y d y = {\frac{1}{2} \int_{0}^{1} d x y^{2}}_{0}^{\sqrt{1 - x^{2}}} = \frac{1}{2} \int_{0}^{1} d x (1 - x^{2}) = {\frac{1}{2} [x - \frac{1}{3} x^{3}]}_{0}^{1} = \frac{1}{3}$

Therefore, both way gives the same results of the integration that is, $\frac{1}{3}$

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$\int_{y = 0}^{1} \int_{x = 0}^{\sqrt{1 - y^{2}}} ydxdy$

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve in an integral way

Calculation of the area under the curve changing boundaries

One App. One Place for Learning.

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∫y=01∫x=01-y2ydxdy

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve in an integral way

Calculation of the area under the curve changing boundaries

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electricity

Force

Physics of Motion

Physical Quantities And Units

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Modern Physics

Study anywhere. Anytime. Across all devices.

$\int_{y = 0}^{1} \int_{x = 0}^{\sqrt{1 - y^{2}}} ydxdy$