Triple Integral in Rectangular Coordinates Converting to Spherical Coordinates

  • Thread starter donald17
  • Start date
  • #1
6
0

Homework Statement



Given that:

24o1zib.jpg


Write an equivalent integral in spherical coordinates.

Homework Equations



bjj38n.jpg


(Triple integral in spherical coordinates.)

r240g7.jpg

2m7gn4g.gif

2ik5iiw.gif

27yu2qw.gif


(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.
 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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 [itex]-\sqrt{a^2- y^2}[/itex] to 0. [itex]x= -\sqrt{a^2- y^2}[/itex] gives [itex]x^2+ y^2= a^2[/itex], 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 [itex]a- \sqrt{a^2- x^2- y^2}[/itex] to a. [itex]z= a- \sqrt{a^2- x^2- y^2}[/itex] gives [itex]x^2+ y^2+ (z-a)^2= a^2[/itex], 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 [itex]x^2+ y^2+ (z-a)^2= a^2[/itex] 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 [itex]\rho[/itex] go from 0 to a, [itex]\theta[/itex] from [itex]\pi[/itex] to [itex]3\pi/2[/itex], and [itex]\phi[/itex] from [itex]\pi/2[/itex] to [itex]\pi[/itex]. The integrand is still "1" and the "differential of volume" is still [itex]\rho sin^2(\phi)d\phi d\theta d\rho[/itex]. 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 [itex](1/8)(4/3)\pi a^3= (1/6)\pi a^3[/itex]. You can use that as a check.
 
  • #3
6
0
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?
 

Related Threads on Triple Integral in Rectangular Coordinates Converting to Spherical Coordinates

Replies
4
Views
2K
Replies
5
Views
979
Replies
4
Views
8K
Replies
17
Views
6K
Replies
3
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Top