can someone express this in polar coordinates please $\displaystyle \int_{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{dxdy}{(x^2+y^2+1)^{3/2}}$
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Here you go: $\displaystyle x^2+y^2=r^2$ $\displaystyle \mathrm{d}x\cdot\mathrm{d}y=r\cdot\mathrm{d}\theta \cdot\mathrm{d}r$
so is this the right one expression ????? $\displaystyle \int_{0}^{\infty }\int_{0}^{\pi/2 }\frac{rd\vartheta dr}{(r^2+1)^{3/2}}$
Originally Posted by kotsos so is this the right one expression ????? $\displaystyle \int_{0}^{\infty }\int_{0}^{\pi/2 }\frac{rd\vartheta dr}{(r^2+1)^{3/2}}$ No, it is not. With $\displaystyle \theta$ going from 0 to $\displaystyle \pi/2$, you are covering only the first quadrant. This would be correct if your original integrals had been from 0 to infinity.
ok ...but what is the right expression to go on on this exercise please ???????
An expression in which the range of $\displaystyle \theta$ spans all four quadrants.
And what is this expression ?Does anyone knows ?
Originally Posted by kotsos And what is this expression ?Does anyone knows ? You have at least two possibilities: $\displaystyle -\pi \le \theta <\pi$ and $\displaystyle 0 \le \theta < 2 \pi$... Marry Christmas from Serbia $\displaystyle \chi$ $\displaystyle \sigma$
Originally Posted by chisigma You have at least two possibilities: $\displaystyle -\pi \le \theta <\pi$ and $\displaystyle 0 \le \theta < 2 \pi$... Marry Christmas from Serbia $\displaystyle \chi$ $\displaystyle \sigma$ Merry Christmas Christian brother from Serbia...!!!!
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