60.7k views
0 votes
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
by
8.2k points

1 Answer

3 votes

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
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories