Is Changing to Polar Coordinates Always Valid for Double Integrals?

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    Integral Paradox
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zetafunction
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An integral paradox ??

let be [tex]\int_{0}^{\infty}xdx \int_{0}^{\infty}ydy[/tex]

changing to polar coordinates we get that the double integral above shoudl be

[tex]2\int_{0}^{\infty}r^{3}dr[/tex]

althoguh they are all divergent , is this true can we ALWAYS make a change of variable to polar coordinates without any ambiguity ??
 


One can always change from rectangular to polar coordinates. However the transformation you gave is incorrect. The coefficient is not 2 but 1/2.
 


am.. thanks a lot

but my question is, the Area of a Circle is NOT equal to the area of an Square [tex]\frac{C}{S}= \pi[/tex]

hence , how could we be completely sure [tex]\iint _{C} f(x,y)dxdy = \iint _{S} f(x,y)dxdy[/tex]
 


Rectangular coordinates don't necessarily trace out rectangles and Polar coordinates don't necessarily trace out Circles in the xy plane. The path they trace out is predetermined by a rule, eg To describe the path of the unit circle in rectangular coordinates we say x^2+y^2 = 1, and the same path could be described in polar coordinates with x= cos t, y= sin t, t varies from 0 to 2pi.

It's your job to change the bounds and integrand of the integral accordingly when change coordinates so that they still sum the same overall function values over the same domain.