17. Let D be the domain between y = x and y = /x. Calculatexy dA as an iterated integral in the order dx dy and dy dx.

Given: ∫∫DxydA , here D be the domain between y=x and y=x . The double integral ∫∫Dxydxdy can be…

Answered: 17. Let D be the domain between y = x…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

17. Let D be the domain between y = x and y = /x. Calculate
xy dA as an iterated integral in the order dx dy and dy dx.

Expert Solution

Step 1

Given:

$\int \int_{D} x y d A$ , here $D$ be the domain between $y = x and y = \sqrt{x} .$

The double integral $\int \int_{D}^{} x y d x d y$ can be evaluated as,

Here, $x$ varies from $y to y^{2}$ and $y$ varies from $\sqrt{x}$ to $x .$

Therefore,

$\int \int_{D}^{} x y d x d y = \int_{\sqrt{x}}^{x} \int_{y}^{y^{2}} x y d x d y = \int_{\sqrt{x}}^{x} y {(\frac{x^{2}}{2})}_{y}^{y^{2}} d y = \int_{\sqrt{x}}^{x} y (\frac{y^{4}}{2} - \frac{y^{2}}{2}) d y = \frac{1}{2} \int_{\sqrt{x}}^{x} (y^{5} - y^{3}) d y = \frac{1}{2} {(\frac{y^{6}}{6} - \frac{y^{4}}{4})}_{\sqrt{x}}^{x} = \frac{1}{2} (\frac{x^{6}}{6} - \frac{x^{4}}{4} - (\frac{x^{3}}{6} - \frac{x^{2}}{4})) = \frac{1}{48} (4 x^{6} - 6 x^{4} - 4 x^{3} + 6 x^{2})$