Setting up an integral (Spherical Coordinates)

In summary, the problem is to integrate a function over the bounded region Q, which is defined by the sphere x²+y²+z²=2 (ρ=sqrt2) and the cylinder x²+y²=1 (ρ=cscφ). The Jacobian for this region is ρ²sinφ and the limits for ρ go from cscφ to sqrt2, while θ goes from 0 to 2pi. The region can be visualized as two parts, the interior of the sphere and the exterior "love handle" of the cylinder. The limits for φ can be found by solving for the angles at which the cylinder and sphere intersect, which are pi/4 and 3pi/4, respectively
  • #1
Tom2
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Homework Statement



To integrate a function (the function itself is not important) over the region Q. Q is bounded by the sphere x²+y²+z²=2 (ρ=sqrt2) and the cylinder x²+y²=1 (ρ=cscφ).

To avoid any confusion, for the coordinates (ρ,φ,θ), θ is essentially the same θ from polar coordinates in 2 dimensions while φ is the angle measured from the +z axis to ρ.

Homework Equations



Jacobian = ρ²sinφ

The Attempt at a Solution



I can see that the limits for ρ go from cscφ to sqrt2.
Also θ should go from 0 to 2pi.
But I'm not sure how to find the limits for φ (the book says it goes from pi/4 to 3pi/4). How is it justified?
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  • #2
What you say about limits depends on the order of integration. You should establish that clearly first... and consider changing it if it makes setting up the limits come out easier.
Given both objects are symmetric under rotation in the x-y plane what you say about [itex]\theta[/itex] is exactly right regardless of the order of integration.

Next you must visualize the region in question. Since the radius of the sphere is greater than the cylinder clearly the cylinder passes through the sphere. This gives us actually two bounded regions. The interior of the sphere also interior to the cylinder and the interior of the sphere exterior to the cylinder. From what you've stated in the problem statement we can't be sure which of these regions you mean however the region inside the cylinder will have to be done in two parts if you use spherical coordinates. However the outer region (the spherical "love handle" around the cylinder) will work nicely in spherical coordinates and by your "I can see.." statement that's the case you're working and your [itex]\rho[/itex] limits are just right for integrating w.r.t. [itex]\varphi[/itex] after integrating w.r.t. [itex]\rho[/itex].

So then the question is what are the final limits for the polar angle. Again as the region of integration is the "love handle" and contains the equator of the sphere (in the x-y plane at [itex]\varphi = \pi/2[/itex]) it is left to solve for the angles at which the cylinder and sphere intersect. Take a cross section containing the z-axis and you'll see a circle intersecting two parallel lines 1 unit each side of the z-axis in the z,r plane. The coordinates of that point are [itex](z,r) = (\sqrt{2}\cos(\varphi),\sqrt{2}\sin(\varphi)) = (z,1)[/itex] so solve that equation for the angle and you get what the book says.
 

1. What are the advantages of using spherical coordinates when setting up an integral?

Spherical coordinates are advantageous for setting up integrals because they are well-suited for problems involving spherical symmetry, such as those in physics and engineering. They also simplify the calculations of integrals involving curved surfaces and volumes, making them more efficient to solve.

2. How do I convert from Cartesian coordinates to spherical coordinates?

To convert from Cartesian coordinates (x, y, z) to spherical coordinates (ρ, θ, φ), you can use the following equations: ρ = √(x² + y² + z²), θ = tan⁻¹(y/x), and φ = cos⁻¹(z/ρ). Keep in mind that ρ represents the distance from the origin, θ represents the angle in the xy-plane, and φ represents the angle from the positive z-axis.

3. What is the range of values for spherical coordinates?

The range of values for spherical coordinates is ρ ≥ 0, 0 ≤ θ ≤ 2π, and 0 ≤ φ ≤ π. This means that ρ represents a distance, θ represents an angle in radians, and φ represents an angle in radians from 0 to 180 degrees.

4. How do I set up an integral in spherical coordinates for a volume or surface area?

To set up an integral in spherical coordinates for a volume, you would use the formula ∫∫∫ f(ρ, θ, φ)ρ²sinφdρdθdφ, where f(ρ, θ, φ) is the function being integrated. To set up an integral for a surface area, you would use the formula ∫∫ f(ρ, θ, φ)ρsinφdθdφ, where f(ρ, θ, φ) is the function representing the surface.

5. How do I choose the limits of integration for a spherical coordinate integral?

The limits of integration for a spherical coordinate integral depend on the shape and boundaries of the region being integrated. Typically, ρ will have a lower limit of 0 and an upper limit corresponding to the radius of the region. θ will have a lower limit of 0 and an upper limit of 2π, representing a full rotation. φ will have a lower limit of 0 and an upper limit corresponding to the angle of the region's boundaries from the positive z-axis.

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