## Related questions with answers

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

Solutions

VerifiedLet $I = \displaystyle \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^{4} xdzdydx$, 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.

The goal of the problem is to write an integral using spherical and cylindrical coordinates. We need to read the boundaries for rectangular coordinates from the integral to do this.

*How do we read boundaries for rectangular coordinates? The limits of the innermost integral refer to which variable?*

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