Triple Integral with Spherical Coordinates

  • #1

Homework Statement


Evaluate [tex]\int\int\int[/tex] 1/[tex]\sqrt{x^{2}+y^{2}+z^{2}+3}[/tex] over boundary B, where B is the ball of radius 2 centered at the origin.


Homework Equations


Using spherical coordinates:
x=psin[tex]\Phi[/tex]cos[tex]\Theta[/tex]
y=psin[tex]\Phi[/tex]sin[tex]\Theta[/tex]
z=pcos[tex]\Phi[/tex]

Integral limits:
dp - [0,2]
d[tex]\Phi[/tex] - [0,[tex]\pi[/tex]]
d[tex]\Theta[/tex] - [0,2[tex]\pi[/tex]]

The Attempt at a Solution


I am just having trouble finding a good substitution for the integrand. When I substitute x,y, and z with the spherical substitutions, I just get a huge jumbled mess that I can't make any sense of.
 

Answers and Replies

  • #2
whoops sorry, the integrand should be
[tex]\frac{dV}{\sqrt{x^{2}+y^{2}+z^{2}+3}}[/tex]
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
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When I substitute x,y, and z with the spherical substitutions, I just get a huge jumbled mess that I can't make any sense of.

Show us. And keep in mind that [itex]\sin^2\eta+\cos^2\eta=1[/itex].
 
  • #4
lol even trying to type that out is a huge jumbled mess in itself. i understand the property you gave me, it's just that i am not able to factor out enough terms such that i leave that identity intact.
 

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