Triple Integral in Rectangular Coordinates Converting to Spherical Coordinates

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SUMMARY

The discussion focuses on converting a triple integral from rectangular coordinates to spherical coordinates for a specific region defined by a sphere of radius 'a' shifted up 'a' units along the z-axis. The limits of integration for theta are established as 0 to π/2, while the integration region is identified as the lower half of the sphere in the third quadrant. The final integral evaluates to (1/6)πa³, representing one-eighth of the sphere's volume. The conversion process emphasizes the importance of correctly identifying the limits for φ and ρ in spherical coordinates.

PREREQUISITES
  • Understanding of spherical coordinates and their conversion from rectangular coordinates.
  • Familiarity with triple integrals and volume calculations in multivariable calculus.
  • Knowledge of the geometric interpretation of spheres and octants in three-dimensional space.
  • Proficiency in using mathematical notation and symbols relevant to calculus.
NEXT STEPS
  • Study the derivation of spherical coordinates and their applications in triple integrals.
  • Learn about volume integrals in spherical coordinates, focusing on different regions of integration.
  • Explore the geometric properties of spheres and their sections in three-dimensional space.
  • Practice converting various integrals from rectangular to spherical coordinates with different limits.
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and integral transformations, as well as professionals in fields requiring spatial analysis and volume calculations.

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Homework Statement



Given that:

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Write an equivalent integral in spherical coordinates.

Homework Equations



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(Triple integral in spherical coordinates.)

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(Conversions from rectangular to spherical coordinates.)(What spherical coordinates entail)

The Attempt at a Solution



The region is going to be the region of the lower part of a sphere with radius a moved up a units on the z-axis that lies in the -x and -y plane. The limits of integration for theta should be 0 to pi/2. Need assistance with limits for phi and rho.
 
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y ranges from -a to 0 so we are talking about the lower half plane in the xy- plane.

For each y, x ranges from -\sqrt{a^2- y^2} to 0. x= -\sqrt{a^2- y^2} gives x^2+ y^2= a^2, the circle with center (0,0) and radius a. Since x is the negative square root and we already know that we are in the lower half plane (y< 0), we are looking at the portion of that circle in the third quadrant.

For each (x,y), z ranges from a- \sqrt{a^2- x^2- y^2} to a. z= a- \sqrt{a^2- x^2- y^2} gives x^2+ y^2+ (z-a)^2= a^2, the sphere with center at (0,0,a) and radius a. Since z goes up to a, we are in the lower half sphere. That is, we are looking at the part of the sphere x^2+ y^2+ (z-a)^2= a^2 where x and y are negative and z< a.

I would be inclined to introduce new coordinates: x'= x, y'= y, z'= z- a. In those coordinates, the region we are integrating over is the "seventh octant" in which x, y, and z are all negative. To get that sphere, let \rho go from 0 to a, \theta from \pi to 3\pi/2, and \phi from \pi/2 to \pi. The integrand is still "1" and the "differential of volume" is still \rho sin^2(\phi)d\phi d\theta d\rho. In fact, because the integrand is "1", this integral is just the volume of the region: 1/8 of the volume of a sphere of radius a and so is (1/8)(4/3)\pi a^3= (1/6)\pi a^3. You can use that as a check.
 
Shouldn't we be integrating the region of the sphere in the -x, -y, +z plane, since z goes from the bottom of the sphere, moved up a on the z-axis, to the plane z=a?
 

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