calcular$\int_Q\frac{1}{|x|} \, dx$ en$Q=[0,1]^2$

Question

Deje$Q=[0,1]^2$, calcule la integral:

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Intenté tomar$$\int_Q\frac{1}{|x|} \, dx$, luego la integral es:

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Luego cambié a las coordenadas polares$x=(x_1,x_2)$, por lo que el integrando es$$\int_Q\frac{1}{\sqrt{x_1^2+x_2^2}} \, dx_1 \, dx_2$.

No pude encontrar el dominio de la integral. Tomé$(x_1,x_2)=(r\cos\theta,r\sin\theta)$ y$\frac{1}{\sqrt{r^2}}r=1$ de$0\leq r\cos\theta\leq1$.

Luego obtuve$0\leq r\sin\theta\leq1$, pero no logré encontrar el intervalo$(x_1,x_2)\in Q$, ¿es$0\leq r\leq\sqrt2$?

El resultado final de WA es:$\theta$

Answer 1

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Answer 2

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$$ \begin{align}I &= \int_0^{\pi/2}\int_0^{\min(1/\cos\theta,1/\sin\theta)} \,\partial r \,\partial \theta\\ &= \int_0^{\pi/4}\int_0^{1/\cos\theta} \, \partial r \, \partial \theta + \int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} \, \partial r \, \partial \theta \\ &= \int_0^{\pi/4}\frac{1}{\cos\theta} \, \partial \theta + \int_{\pi/4}^{\pi/2} \frac{1}{\sin\theta} \, \partial \theta \\ &= \log(\tan\theta+\sec\theta)|_0^{\pi/4} -\log(\cot\theta+\csc\theta)|_{\pi/4}^{\pi/2}\\ &= \log(1+\sqrt 2) - 0-0 + \log(1+\sqrt{2}) \\ &= 2\log(1+\sqrt 2)\end {align} $$

Tenga en cuenta que utilicé el hecho de que$$0 \leq \tan\theta \leq \infty \implies \theta \in [0,\pi/2]$ y$$0 \leq r\cos\theta \leq 1, 0 \leq r\sin\theta \leq 1 \implies 0 \leq r \leq \min(1/\cos\theta,1/\sin\theta) $.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\bracks{0,1}^{2}}{\dd x\,\dd y \over \root{x^{2} + y^{2}}} & = \int_{\bracks{0,1}^{2}}{\dd x\,\dd y \over r} = \int_{\bracks{0,1}^{2}}\bracks{\partiald{\pars{x/r}}{x} - \partiald{\pars{-y/r}}{y}}\,\dd x\,\dd y \\[5mm] & = \int_{\bracks{0,1}^{2}}\bracks{% \nabla\times\pars{-\,{y \over r}\,\hat{x}\ +\ {x \over r}\,\hat{y}}}_{z} \,\dd x\,\dd y \\[5mm] & = \int_{\partial\bracks{0,1}^{2}}{-y\,\dd x + x\,\dd y \over \root{x^{2} + y^{2}}} \qquad\pars{~Stokes\ Theorem~} \\[5mm] & = \int_{0}^{1}{\dd y \over \root{1 + y^{2}}} + \int_{1}^{0}{-\dd x \over \root{x^{2} + 1}} = 2\int_{0}^{1}{\dd x \over \root{x^{2} + 1}} \\[5mm] \stackrel{x\ \mapsto\ \tan\pars{x}}{=}\,\,\,& 2\int_{0}^{\pi/4}\sec\pars{x}\,\dd x = 2\bracks{\vphantom{\Large A}\ln\pars{\sec\pars{x} + \tan\pars{x}}}_{\ 0}^{\ \pi/4} \\[5mm] &= 2\ln\pars{\sec\pars{\pi \over 4} + \tan\pars{\pi \over 4}} = \bbx{2\ln\pars{\root{2} + 1}} \approx 1.7627 \end{align}