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Mathematica · Feb 3, 2009, 09:44 PM

Please show the answer step by step... I know that it take around 2 pages but I would really appreciated if you show me the entire steps... thanks .

Convert the integral from rectangular coordinates to spherical coordinates.
it is a triple integral.. from left to right

Integ (-2 to 2), integ (sqrt [4-x^2] to -sqrt[4-x^2]), iteg (x^2+y^2) to 4)) integrant X dzdydx

again... I have the answer but I don't know how to get it...
thank you .

galactus · Feb 4, 2009, 07:55 AM

Is this your integral:

$\int_{-2}^{2} \;\ \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \;\ \int_{x^{2}+y^{2}}^{4}x \;\ dzdydx$

If so, note that in spherical coordinates $x={\rho}sin({\phi})cos({\theta})$

$y={\rho}sin({\phi})sin({\theta})$

$z={\rho}cos({\phi})$

Try to see what this represents. The region in the xy plane is a circle centered at the origin with radius 2.

Mathematica · Feb 4, 2009, 12:33 PM

[QUOTE=galactus;1527171]Is this your integral:

$\int_{-2}^{2} \;\ \int_{\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \;\ \int_{x^{2}+y^{2}}^{4}x \;\ dzdydx$

If so, note that in spherical coordinates $x={\rho}sin({\phi})cos({\theta})$

$y={\rho}sin({\phi})sin({\theta})$

$z={\rho}cos({\phi})$

Try to see what this represents. The region in the xy plane is a circle centered at the origin with radius 2.[/QUOT]

the middle integral is not correct , the lower limit has to be negative -sqrt(4-x^2)

galactus · Feb 4, 2009, 02:54 PM

That's just a typo. Thanks for the catch.