Convert the integral from rectangular coordinates to both cylindrical and spherical coordinates, and evaluate the simplest iterated integral. ∫_(-2)^2 ∫_(-√4-x²)^(√4-x²) ∫_(x²+y²)^4 x dz dy dx

Let $I = \displaystyle \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^{4} xdzdydx$, then in cylindrical coordinates, $I = \displaystyle \int_{0}^{2}\int_{0}^{2\pi}\int_{r^2}^{4} r\cos\theta\cdot r dzd\theta dr$ and in spherical coordinates,

$I =\displaystyle\int_{0}^{2\pi}\int_{0}^{\arctan\frac{1}{2}}\int_{0}^{4\sec \phi} \rho\sin\phi\cos\theta\cdot \rho^2\sin\phi d\rho d\phi d\theta + \int_{0}^{2\pi}\int_{\arctan\frac{1}{2}}^{\frac{\pi}{2}}\int_{0}^{\cot \phi \csc \phi} \rho\sin\phi\cos\theta\cdot \rho^2\sin\phi d\rho d\phi d\theta$

Note: The reason we need to split into two triple integrals for the spherical coordinates is because $\rho$ is originated from the origin and terminates at the surface. However, we have two different surfaces here. One surface is the paraboloid with equation $z = x^2 + y^2$ or $\rho = \cot \phi \csc \phi$ in spherical coordinates (see the red dotted line on the right) and the other surface is the plane with equation $z =4$ or $\rho = 4\sec \phi$ in spherical coordinates (see the blue dotted line on the right).

In cylindrical coordinates, $x = r\cos\theta$, $y=r\sin\theta$ and $z = z$. Furthermore, $x^2 + y^2 = r^2$ and the Jacobian is $r$.

In spherical coordinates, $x = \rho \sin \phi \cos\theta$, $y=\rho \sin \phi \sin\theta$ and $z = \rho \cos \phi$. Furthermore, $x^2 + y^2 +z^2= \rho^2$ and the Jacobian is $\rho^2 \sin \phi$. Notice that $x^2 + y^2 +z^2= \rho^2 \rightarrow x^2 + y^2 +(\rho \cos \phi)^2 = \rho^2 \rightarrow x^2 + y^2 = \rho^2 - \rho^2 \cos^2 \phi \rightarrow \rightarrow x^2 + y^2 = \rho^2 (1- \cos^2 \phi ) \rightarrow x^2 + y^2 = \rho^2\sin^2 \phi$.

Also here, the solid represented by the three sets of limits is the region bounded above by the plane $z = 4$ and bounded below by the paraboloid $z = x^2 + y^2$ and the region represented by the outer two sets of limits (for $dydx$) is the disk centered at origin with radius 2.