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Set up the following integral in spherical coordinates, compute it:

∭ᵣ z/(x²+y²+z²)³/2 d V
where R is the region in the first octant defined by x² + y² + z² ≤ 9 with z ≥ 2/3, set up the integral in cylindrical coordinates. Do not compute the integral.

User LeviX
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Final answer:

In cylindrical coordinates, the integral of z/(x²+y²+z²)³/2 over the specified region is transformed by substituting r² for x²+y² and including the Jacobian (r) for volume integration.

Step-by-step explanation:

The integral ∫∫∫R z/(x²+y²+z²)³/2 dV, where R is the region in the first octant defined by x² + y² + z² ≤ 9 with z ≥ 2/3, can be expressed in cylindrical coordinates. In cylindrical coordinates, we denote z as z, r as the distance from the origin to the projection of the point on the xy-plane, and θ as the angle from the positive x-axis to the projection. Since the region is in the first octant, θ ranges from 0 to π/2. The integral becomes:

∫0∫2/3∫0 z/(r²+z²)³/2 ⋅ r dz dr dθ

Here, the outer integral varies over θ from 0 to π/2, the middle integral over z from 2/3 to √(9 - r²), and the inner integral over r from 0 to 3. Notice that the function has been adjusted by substituting r² for x² + y², and including the Jacobian (r) for the transformation from Cartesian to cylindrical coordinates when integrating with respect to the volume element dV.

User RicLeal
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