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I'm not quite sure how to complete this question

$D$ is the region bounded by $x = 0, y=0,x+y=1$, and $x+y=4$. Using the change of variables $x = u -uv, y = uv$ and the Jacoian, evaluate the double integral $$\iint_D \frac {dxdy}{x+y}$$I found the Jacobian $$\begin{vmatrix}1-v&-u\\v&u\\ \end{vmatrix} = u$$ and after substituting $x,y$ for $u,v$ I got $$\iint_{D^*} u\frac {dudv}{u -uv+uv} =\iint_{D^*} dudv$$ but I have no idea how to find $D^*$

Any help would be appreciated.

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  • $\begingroup$ Anyone.....? :( $\endgroup$ – user3128875 Apr 29 '14 at 7:11
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Let us first carry out the integral given in (x,y). $$ \int_0^1 \left( \int_{1-x}^{4-x} \frac{1}{x+y} dy \right) dx + \int_1^4 \left( \int_{0}^{4-x} \frac{1}{x+y} dy \right) dx = 3. $$ This integral is bounded by the lines x=0, y=0, y=4-x and y=1-x. To be able to find the limits of the integral given in (u,v), we need to find what these lines become in the new coordinate system (v,u). First, note that $v=y/(x+y)$ and $u=x+y$. $x=0$ means $v=1$. As for $y=0$, it corresponds to $v=0$. Similarly, $y=1-x$ and $y=4-x$ correspond to $u=4$ and $u=1$, respectively. Your new integration domain is now a rectangle in the (v,u) coordinate system, which simplifies the limits of the integrals.

The result is $$ \int_1^4 \left( \int_0^1 u \frac{1}{u-uv+uv} dv \right) du=\int_1^4 \left( \int_0^1 dv \right) du=3. $$

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