Monday Math 148

Find $I=\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx$ .

One could integrate over y by using
$\frac{x^5-y^3}{x^2+y^2}=\frac{x^5}{x^2+y^2}-\frac{y^3}{x^2+y^2}=\frac{x^5}{x^2+y^2}-y+\frac{x^2y}{x^2+y^2}$ ; however, when you put in the limits, the resulting integral over x becomes very difficult.

Note that the region of integration, the region bounded by the parabolas $y=\frac{x^2}2+x-\frac12$ and $y=-\frac{x^2}2+x+\frac12$ , has twofold rotational symmetry about the origin. Thus, we consider the change in variables $u=-x,\;\;v=-y$ , which represents axes rotated by 180°. We then have
$\begin{eqnarray}I&=&\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx\\&=&\int_{1}^{-1}\int_{-\frac{u^2}2+u+\frac12}^{\frac{u^2}2+u-\frac12}\frac{(-u)^5-(-v)^3}{(-u)^2+(-v)^2}\,(-dv)\,(-du)\\&=&\int_{-1}^{1}\int_{\frac{u^2}2+u-\frac12}^{-\frac{u^2}2+u+\frac12}\frac{-u^5+v^3}{u^2+v^2}\,dv\,du\\I&=&-\int_{-1}^{1}\int_{\frac{u^2}2+u-\frac12}^{-\frac{u^2}2+u+\frac12}\frac{u^5-v^3}{u^2+v^2}\,dv\,du\end{eqnarray}$ ,
and if we rename u and v as x and y, respectively, we see
$\begin{eqnarray}I&=&-\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx\\I&=&-I\\I&=&{0}\end{eqnarray}$ ; for our integrand $f(x,y)=\frac{x^5-y^3}{x^2+y^2}$ , we have $f(-x,-y)=-f(x,y)$ , which combined with the rotational symmetry of the region of integration, means that the integral is zero.

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Tags: Double Integral, Integral, Math, Monday Math, symmetry

This entry was posted on December 27, 2010 at 12:34 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Monday Math 148

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