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I am supposed to evaluate a double integral using change of variables. The integral is: $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}dydx $

I am given these equations to switch between variables:

$ u=\frac{y}{x}$ and $v=x+y$

Calculations:

The first thing to calculate is the Jacobian:

$dxdy$ = $\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{bmatrix}$

$\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}=1$

$\frac{\partial u}{\partial x}=\frac{-y}{x^{2}}$ and $\frac{\partial u}{\partial y}=\frac{1}{x}$

Therefore the Jacobian = $dxdy$ = $\frac{-y}{x^{2}}-\frac{1}{x}$

= $\frac{-y-x}{x^{2}}$

and the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}(\frac{-y-x}{x^{2}}) dudv $

Since $v=x+y$,

the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(v)e^{(v)}(-v)}{{x^{2}}}dudv $

$ \int_{0}^{1}\int_{0}^{x}\frac{(-v^{2})e^{(v)}}{{x^{2}}}dudv $

At this point I am confused how to substitute for the x in the integral bound and the denominator as well as changing the boundaries in the integrals. I know I want the entire integral in terms of u and v. I was thinking about using $x=v-y$ but that leaves it in terms I don't want.

The answer is $e^{2}-e-1$

Thanks for the help.

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  • $\begingroup$ You need to solve your current variables first in terms of $u$ and $v$ and then calculate the Jacobian of $\frac{\delta x}{\delta u}$, etc. $\endgroup$ – Brent Mar 21 '15 at 2:27
  • $\begingroup$ Brent, I already calculated the Jacobian. I don't see how to solve for x without getting a y back. $\endgroup$ – Chan Hunt Mar 21 '15 at 2:58
  • $\begingroup$ In addition to the above point, I guess it makes more sense to leave $x$ untouched (or let $v=x$) than $v=x+y$. $\endgroup$ – Fan Zheng Mar 21 '15 at 2:59
  • $\begingroup$ In you COV $y=ux$, so $v=x+ux=(1+u)x$, so $x=v/(1+u)$. $\endgroup$ – Fan Zheng Mar 21 '15 at 3:00
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1.) Transform differentials:

$$\begin{gathered} (x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx \hfill \\ u = \frac{y}{x},v = x + y \hfill \\ du = - \frac{y}{{{x^2}}}dx + \frac{1}{x}dy \hfill \\ dv = dx + dy \hfill \\ du \wedge dv = - \frac{y}{{{x^2}}}dx \wedge dy + \frac{1}{x}dy \wedge dx \hfill \\ du \wedge dv = - \frac{{x + y}}{{{x^2}}}dx \wedge dy = \frac{{x + y}}{{{x^2}}}dy \wedge dx \hfill \\ \frac{1}{v}du \wedge dv = \frac{1}{{{x^2}}}dy \wedge dx \hfill \\ \frac{1}{{{x^2}}}dydx = \frac{1}{v}dudv \hfill \\ (x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = {e^v}dudv \hfill \\ \end{gathered}$$

2.) Find inverse function:

$$\begin{gathered} u = \frac{y}{x},v = x + y \Rightarrow uv - y = x\frac{{{y^2}}}{{{x^2}}} = x \cdot {u^2} \hfill \\ \Rightarrow uv - x - y = x \cdot {u^2} - x \hfill \\ \Rightarrow uv - v = x \cdot ({u^2} - 1) \hfill \\ \Rightarrow x = \frac{{v(u - 1)}}{{{u^2} - 1}} = \frac{v}{{u + 1}} \hfill \\ v = x + y \Rightarrow \hfill \\ y = v - x = v - \frac{v}{{u + 1}} = v(1 - \frac{1}{{u + 1}}) = v \cdot \frac{u}{{u + 1}} \hfill \\ x = v \cdot \frac{1}{{u + 1}} \hfill \\ y = v \cdot u\frac{1}{{u + 1}} \hfill \\ \end{gathered} $$

3.) Transform boundaries:

$$\begin{gathered} x = 0 \Rightarrow v = 0 \hfill \\ x = 1 \Rightarrow v = u + 1 \hfill \\ y = 0 \Rightarrow v \cdot u = 0 \Rightarrow u = 0 \vee v = 0 \hfill \\ y = x \Rightarrow u = 1 \hfill \\ \end{gathered}$$

4.) Integrate:

$$\begin{gathered} \int\limits_0^1 {\int\limits_0^x {(x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = \int\limits_0^1 {\int\limits_0^{u + 1} {{e^v}dvdu} } } } = \int\limits_0^1 {({e^{u + 1}} - 1)du} \hfill \\ \int\limits_0^1 {({e^{u + 1}} - 1)du} = {e^2} - e - 1 \hfill \\ \end{gathered}$$

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$ \int_{0}^{1}\int_{0}^{x}\frac{(-v^{2})e^{(v)}}{{x^{2}}}dudv $

$ y=u x, v=x + u x= (1+u)x, x=v/(1+u) , $ plug in for x

$ \int_{0}^{1}\int_{0}^{x} -e^v (1+u^2) du dv $

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