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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


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Let I=224x24x2x2+y24xdzdydxI = \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=0202πr24rcosθrdzdθdrI = \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=02π0arctan1204secϕρsinϕcosθρ2sinϕdρdϕdθ+02πarctan12π20cotϕcscϕρsinϕcosθρ2sinϕdρdϕdθI =\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 ρ\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=x2+y2z = x^2 + y^2 or ρ=cotϕcscϕ\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=4z =4 or ρ=4secϕ\rho = 4\sec \phi in spherical coordinates (see the blue dotted line on the right).

In cylindrical coordinates, x=rcosθx = r\cos\theta, y=rsinθy=r\sin\theta and z=zz = z. Furthermore, x2+y2=r2x^2 + y^2 = r^2 and the Jacobian is rr.

In spherical coordinates, x=ρsinϕcosθx = \rho \sin \phi \cos\theta, y=ρsinϕsinθy=\rho \sin \phi \sin\theta and z=ρcosϕz = \rho \cos \phi. Furthermore, x2+y2+z2=ρ2x^2 + y^2 +z^2= \rho^2 and the Jacobian is ρ2sinϕ\rho^2 \sin \phi. Notice that x2+y2+z2=ρ2x2+y2+(ρcosϕ)2=ρ2x2+y2=ρ2ρ2cos2ϕx2+y2=ρ2(1cos2ϕ)x2+y2=ρ2sin2ϕx^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=4z = 4 and bounded below by the paraboloid z=x2+y2z = x^2 + y^2 and the region represented by the outer two sets of limits (for dydxdydx) is the disk centered at origin with radius 2.

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