Chapter 5: Q27P (page 248)

$\int_{x = 0}^{4} \int_{y = 0}^{\sqrt{x}} \sqrt[y]{x} dydx$

Short Answer

Expert verified

The required solution is $\frac{32}{5}$

Step by step solution

Definition of double integral

The double integral of f(x,y) over the areaA in 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:

$l = \int_{0}^{4} d x \sqrt{x} \int_{0}^{\sqrt{x}} y d y = {\int_{0}^{4} d x \sqrt{x \frac{1}{2} y^{2}}}_{0}^{\sqrt{x}} = \frac{1}{2} \int_{0}^{4} x^{3 / 2} d x = {\frac{1}{5} x^{5 / 2}}_{0}^{4} = \frac{4^{5 / 2}}{5} = \frac{32}{5}$

Calculation of the area under the curve changing boundaries

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

$y : 0 \to 2, and x : y^{2} \to 4$

So, the integration

$l = \int_{0}^{2} y d y \int_{y^{2}}^{4} \sqrt{x} d x = {\int_{0}^{2} y d y \frac{2}{3} x^{3 / 2}}_{y}^{4} = \frac{2}{3} \int_{0}^{2} y d y (8 - y^{3}) = {\frac{2}{3} \int_{0}^{2} [4 y^{2} - \frac{1}{5}]}_{0}^{2} = \frac{32}{5}$

Therefore, both way gives the same results of the integration that is, $\frac{32}{5}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

$\int_{x = 0}^{4} \int_{y = 0}^{\sqrt{x}} \sqrt[y]{x} dydx$

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

Space Physics

Physical Quantities And Units

Torque and Rotational Motion

Circular Motion and Gravitation

Oscillations

Linear Momentum

Study anywhere. Anytime. Across all devices.

∫x=04∫y=0xxydydx

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

Space Physics

Physical Quantities And Units

Torque and Rotational Motion

Circular Motion and Gravitation

Oscillations

Linear Momentum

Study anywhere. Anytime. Across all devices.

$\int_{x = 0}^{4} \int_{y = 0}^{\sqrt{x}} \sqrt[y]{x} dydx$