Q 54 - Calculus

Question

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.

$\int {\int }_R\frac{dA}{x^2+2xy+y^2},$

where $R=\left\{\left(x,y\right)|1\le x\le 2 and 0\le y\le 1\right\}$.

Step-by-Step Solution

Answer

The value of given double integral is :-

$\int {\int }_R\frac{dA}{x^2+2xy+y^2}=\log \left(\frac{4}{3}\right)$

where $R=\left\{\left(x,y\right)|1\le x\le 2 and 0\le y\le 1\right\}$

Step 1. Given Information

We have given the following double integral :-

$\int {\int }_R\frac{dA}{x^2+2xy+y^2},$

where $R=\left\{\left(x,y\right)|1\le x\le 2 and 0\le y\le 1\right\}$

We have to evaluate this double integral.

Step 2. Use iterated integrals

The given double integral is :-

$\int {\int }_R\frac{dA}{x^2+2xy+y^2},$

where $R=\left\{\left(x,y\right)|1\le x\le 2 and 0\le y\le 1\right\}$

In order to solve this double integral we will firstly integrated with $y$.

Then by using Fubini's Theorem, we can writ this double integral as following :-

$\int {\int }_R\frac{dA}{x^2+2xy+y^2}={\int }_1^2{\int }_0^1\frac{1}{x^2+2xy+y^2}dydx$

Then by using iterated integrals, we have :-

${\int }_1^2{\int }_0^1\frac{1}{x^2+2xy+y^2}dydx={\int }_1^2\left({\int }_0^1\frac{1}{x^2+2xy+y^2}dy\right)dx$

Now we can solve this integral as following :-

$$\begin{gathered} {\int }_1^2\left({\int }_0^1\frac{1}{x^2+2xy+y^2}dy\right)dx \\ ={\int }_1^2\left({\int }_0^1\frac{1}{{\left(x+y\right)}^2}dy\right)dx \\ ={\int }_1^2\left({\int }_0^1{\left(x+y\right)}^{-2}dy\right)dx \\ ={\int }_1^2{{\left[-{\left(x+y\right)}^{-1}\right]}^1}_{0}dx \\ ={\int }_1^2-{{\left[\frac{1}{x+y}\right]}^1}_{0}dx \\ ={\int }_1^2-\left[\frac{1}{x+1}-\frac{1}{x}\right]dx \\ ={\int }_1^2\left[\frac{1}{x}-\frac{1}{x+1}\right]dx \end{gathered}$$

$$\begin{gathered} ={\left[\log \left(x\right)-\log \left(x+1\right)\right]}_1^2 \\ ={\left[\log \left(\frac{x}{x+1}\right)\right]}_1^2 \\ =\left[\log \left(\frac{2}{3}\right)-\log \left(\frac{1}{2}\right)\right] \\ =\log \left(\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{1}{2}}}\right) \\ =\log \left(\frac{2}{3}\times \frac{2}{1}\right) \\ =\log \left(\frac{4}{3}\right) \end{gathered}$$