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 .

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.

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.

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

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