Volume of a Sphere (example continued)

 

 

The second integration becomes

 

 

                                            θ = π/2      w = 0       

                              V   =   8    ∫                ∫      w 1/2  (- ½)  dw dθ

                                            θ = 0         w = R2                   

 

 

                                              θ = π/2                    0       

                              V   =   - 4    ∫         (2/3) w 3/2 |      dθ

                                              θ = 0                       R2                   

 

 

 

 

Finally integrate on the variable  θ 

 

 

                                              θ = π/2                          

                              V   =   - 4    ∫         (- 2/3) R3     dθ

                                              θ = 0                                          

 

 

which gives         V = 4 πR3 / 3                                     (result)

 

 

 

Next evaluate the volume of the sphere using spherical coordinates.

 

You will see that the use of spherical coordinates simplifies determining the limits

of integration.

 

Click here to continue.

 



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