Answered: ху dr dy (x² + y²)'

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

Evaluating integrals Evaluate the following integral as they are written.

Expert Solution

Step 1: Given

The integral, $\int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y$ .

Step 2: To determine

To evaluate the integral, $\int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y$ .

Let, $\int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y - - - - (1)$ .

First integrating the above integral with respect to $x$ and keeping $y$ constant, we get

$\int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y = \int_{1}^{2} y (\int_{1}^{4} \frac{x}{{(x^{2} + y^{2})}^{2}} d x) d y$ .

Now, substituting $u = x^{2} + y^{2} \Rightarrow d u = 2 x d x$ . Implies when $x = 1 \Rightarrow u = y^{2} + 1$ and when $x = 4 \Rightarrow u = y^{2} + 16$ .

$\begin{array}{rcl} \int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y & = & \int_{1}^{2} y (\int_{y^{2} + 1}^{y^{2} + 16} \frac{1}{2 u^{2}} d x) d y \end{array}$

Step 3

Therefore,

$\begin{array}{rcl} \int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y & = & \int_{1}^{2} y (\int_{y^{2} + 1}^{y^{2} + 16} \frac{1}{2 u^{2}} d x) d y \\ = & \int_{1}^{2} \frac{y}{2} (\int_{y^{2} + 1}^{y^{2} + 16} \frac{1}{u^{2}} d u) d y \\ = & \int_{1}^{2} \frac{y}{2} {[- \frac{1}{u}]}_{y^{2} + 1}^{y^{2} + 16} d y - - - - \{∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C\} \\ = & \int_{1}^{2} \frac{y}{2} {[- \frac{1}{y^{2} + 16} + \frac{1}{y^{2} + 1}]}_{y^{2} + 1}^{y^{2} + 16} d y \\ = & \int_{1}^{2} \frac{y}{2} [\frac{15}{(y^{2} + 1) (y^{2} + 16)}] d y \\ = & \frac{15}{2} \int_{1}^{2} [\frac{y}{(y^{2} + 1) (y^{2} + 16)}] d y - - - - (2) \end{array}$

Step 4

Thus, integrating equation (2) with respect to $y$ , we get

$\begin{array}{rcl} \int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y & = & \frac{15}{2} \int_{1}^{2} [\frac{y}{(y^{2} + 1) (y^{2} + 16)}] d y \end{array}$ .

Now taking partial fraction of $\begin{array}{rcl} \frac{y}{(y^{2} + 1) (y^{2} + 16)} & = & \frac{y}{15 (y^{2} + 1)} - \frac{y}{15 (y^{2} + 16)} \end{array}$ .

$\begin{array}{rcl} \Rightarrow & \int_{1}^{2} \int_{1}^{4} \frac{x y}{{(x^{2} + y^{2})}^{2}} d x d y = \frac{15}{2} [\int_{1}^{2} (\frac{y}{15 (y^{2} + 1)}) d y - \int_{1}^{2} (\frac{y}{15 (y^{2} + 16)}) d y] \end{array}$ .

Next, substitute $u = y^{2} + 1 \Rightarrow d u = 2 y d y$ in $\begin{array}{rcl} \int_{1}^{2} (\frac{y}{15 (y^{2} + 1)}) d y \end{array}$ .

Such that when $y = 1 \Rightarrow u = 2$ and when $y = 2 \Rightarrow u = 5$ .

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