# Thread: expression in polar coordinates

1. ## expression in polar coordinates

can someone express this in polar coordinates please

$\int_{-\infty }^{+\infty }\int_{-\infty }^{+\infty }\frac{dxdy}{(x^2+y^2+1)^{3/2}}$

2. ## Re: expression in polar coordinates

Here you go:

$x^2+y^2=r^2$

$\mathrm{d}x\cdot\mathrm{d}y=r\cdot\mathrm{d}\theta \cdot\mathrm{d}r$

3. ## Re: expression in polar coordinates

so is this the right one expression ?????

$\int_{0}^{\infty }\int_{0}^{\pi/2 }\frac{rd\vartheta dr}{(r^2+1)^{3/2}}$

4. ## Re: expression in polar coordinates

so is this the right one expression ?????

$\int_{0}^{\infty }\int_{0}^{\pi/2 }\frac{rd\vartheta dr}{(r^2+1)^{3/2}}$

5. ## Re: expression in polar coordinates

Originally Posted by kotsos
so is this the right one expression ?????

$\int_{0}^{\infty }\int_{0}^{\pi/2 }\frac{rd\vartheta dr}{(r^2+1)^{3/2}}$
No, it is not. With $\theta$ going from 0 to $\pi/2$, you are covering only the first quadrant. This would be correct if your original integrals had been from 0 to infinity.

6. ## Re: expression in polar coordinates

ok ...but what is the right expression to go on on this exercise please ???????

7. ## Re: expression in polar coordinates

An expression in which the range of $\theta$ spans all four quadrants.

8. ## Re: expression in polar coordinates

And what is this expression ?Does anyone knows ?

9. ## Re: expression in polar coordinates

Originally Posted by kotsos
And what is this expression ?Does anyone knows ?
You have at least two possibilities: $-\pi \le \theta <\pi$ and $0 \le \theta < 2 \pi$...

Marry Christmas from Serbia

$\chi$ $\sigma$

10. ## Re: expression in polar coordinates

Originally Posted by chisigma
You have at least two possibilities: $-\pi \le \theta <\pi$ and $0 \le \theta < 2 \pi$...

Marry Christmas from Serbia

$\chi$ $\sigma$
Merry Christmas Christian brother from Serbia...!!!!