Converting triple integral to spherical

In summary, the conversation is about evaluating an integral using spherical coordinates and determining the correct limits of integration. The first integral traces out a semicircle in the x-z plane and rotates along the y-axis, while the second integral represents a sphere with radius 1. The transformed integration region is a hemisphere in the positive x-direction. The limits for rho, phi, and theta are 0 to 1, 0 to pi, and 0 to pi respectively, assuming the x-axis aligns with phi = pi/2.
  • #1
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  • #2
Well the best way of checking the limits in a case like this is to draw the regions you are integrating over (ignoring the integrand).

What is the integration region of the first integral?

I'll help you with this one: x goes from 0 to 1, z goes from -sqrt(1-x^2) to sqrt(1+x^2) so these two integrals trace out a semicircle in the x-z plane (in the x>0 region). Since y goes from -sqrt(1-x^2-z^2) to sqrt(1-x^2-z^2) this rotates the semicircle along the y-axis to give a...


What is the second (transformed) integration region?

Are these the same? If so good, if not change the limits so they are.
 
  • #3
I don't really get what you are saying.

I know that sqrt(1-x^2-z^2) and -sqrt(1-x^2-z^2) makes a sphere with radius 1, which can be turned into rho=1.

I just can't figure out what phi and theta run to. Since I think it's a sphere (right?) can I assume phi and theta run complete cycles? Is that right?

Can someone please tell me these two limits (and if rho does indeed =1) so that I can see if I'm doing this right?
 
  • #4
Wait, if it's a sphere of rho=1, it then projects a shadow on the x-z plane of a circle with radius one, but the only area of interest is from x=0 to x=1.

So a semi-sphere that exists only in the positive-x?

Is that the shape and the correspondingly correct limits?
 
  • #5
hemi-sphere in positive x sounds good to me, now what do your spherical limits become...
 
  • #6
rho from 0 to 1

phi from 0 to pi

and theta from 0 to pi

Right?
 
  • #7
Right!?
 
  • #8
depends on your convention if phi is teh angle around teh horizontal & the x-axis aligns with phi = pi/2, which i think it does, then yes
 

1. What is the formula for converting a triple integral to spherical coordinates?

The formula for converting a triple integral to spherical coordinates is: dV = ρ² sinφ dρ dφ dθ, where ρ represents the radial distance, φ represents the polar angle, and θ represents the azimuthal angle.

2. How do I determine the limits of integration in spherical coordinates?

The limits of integration in spherical coordinates depend on the shape of the region being integrated. In general, the radial distance ρ ranges from 0 to the maximum value of the radius, the polar angle φ ranges from 0 to π, and the azimuthal angle θ ranges from 0 to 2π.

3. What are the advantages of using spherical coordinates in triple integrals?

Spherical coordinates are particularly useful for integrating over regions with spherical symmetry, such as spheres, cones, and spheres with cylindrical holes. They also simplify the evaluation of integrals involving spherical objects or fields, such as gravitational or electromagnetic fields.

4. Can I convert any triple integral to spherical coordinates?

No, not every triple integral can be converted to spherical coordinates. It is important to consider the symmetry of the region being integrated and whether spherical coordinates are a suitable choice for the problem at hand.

5. How do I convert a triple integral to spherical coordinates in practice?

To convert a triple integral to spherical coordinates, start by identifying the limits of integration for each variable and substituting them into the formula dV = ρ² sinφ dρ dφ dθ. Then, substitute the appropriate expressions for ρ, φ, and θ in terms of x, y, and z. Finally, evaluate the integral using the new limits and variable substitutions.

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