Question

Alguém pode me ajudar?

Answer 1

$I_{R}=\displaystyle\iint_{\Omega}{e^{-x^{2}-y^{2}}\,dx\,dy}\\ \\ \\ I_{R}=\iint_{\Omega}{e^{-(x^{2}+y^{2})}\,dx\,dy}$

sendo $\Omega$ o disco dado por

$x^{2}+y^{2}\leq R^{2}.$

(disco com centro na origem e raio $R.$)

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Mudança para coordenadas polares:

$\begin{array}{cc} \left\{ \begin{array}{l} x=r\cos \theta\\ \\ y=r\,\mathrm{sen\,}\theta \end{array} \right.~~&~~\begin{array}{c} 0\leq r\leq R\\ \\ 0\leq \theta \leq 2\pi \end{array} \end{array}$

O módulo do Jacobiano desta transformação é $|\mathrm{Jac\,}\phi|=r.$
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A função integranda $f(x,\;y)=e^{-(x^{2}+y^{2})}$ sofrerá transformação:

$f(x,\;y)=g(r,\;\theta)\\ \\ e^{-(x^{2}+y^{2})}=e^{-((r\cos \theta)^{2}+(r\,\mathrm{sen\,}\theta)^{2})}\\ \\ e^{-(x^{2}+y^{2})}=e^{-r^{2}}~\Rightarrow~\boxed{\begin{array}{c} g(r,\;\theta)=e^{-r^{2}} \end{array}}$
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Escrevendo a integral iterada em coordenadas polares:

$I_{R}=\displaystyle\iint_{\Omega}{f(x,\;y)\,dx\,dy}\\ \\ \\ =\iint_{\Omega_{r,\theta}}{g(r,\;\theta)\cdot |\mathrm{Jac\,}\phi|\,dr\,d\theta}\\ \\ \\ =\iint_{\Omega_{r,\theta}}{e^{-r^{2}}\cdot r\,dr\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{R}{e^{-r^{2}}\cdot r\,dr\,d\theta}$

$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{R}{-\dfrac{1}{2}\,e^{-r^{2}}\cdot (-2r)\,dr\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left.\left(-\dfrac{1}{2}\,e^{-r^{2}} \right )\right|_{0}^{R}d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left(-\dfrac{1}{2}\,e^{-R^{2}}+\dfrac{1}{2}\,e^{-0^{2}} \right )d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left(-\dfrac{1}{2}\,e^{-R^{2}}+\dfrac{1}{2} \right )d\theta}$

$=\left(\dfrac{1}{2}-\dfrac{1}{2}\,e^{-R^{2}} \right )\displaystyle\int\limits_{0}^{2\pi}{d\theta}\\ \\ \\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\,e^{-R^{2}} \right )\cdot \theta|_{0}^{2\pi}\\ \\ \\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\,e^{-R^{2}} \right )\cdot (2\pi-0)\\ \\ \\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\,e^{-R^{2}} \right )\cdot 2\pi\\ \\ \\ =\pi\,(1-e^{-R^{2}})$