Write iterated Integrals equal to the triple integral integral integral_R integr

Question

Write iterated Integrals equal to the triple integral integral integral_R integral 1/x^2 + y^2 + z^2 dV where R is region lying inside the sphere of radius 4 about the origin and above the cone z = Squareroot x^2 y^2 using each of rectangular, cylindrical and spherical coordinates. Briefly explain which you would prefer to evaluate and why. (In addition to point awarded for correctly giving iterated Integrals in each coordinate system, points will be awarded for the soundness of your preference and ryour eason for it.) You do not need to evaluate any of the interated integrals.

Explanation / Answer

given sphere x2+y2+z2=42 => z =(42-(x2+y2))=(16-x2-y2)

cone z=(x2+y2)

(42-(x2+y2))=(x2+y2)

(42-(x2+y2))=(x2+y2)

2(x2+y2)=16

(x2+y2)=8

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=>-22 <=x <=22 , -(8-x2)<=y<=(8-x2),(x2+y2) <=z<=(16-x2-y2)

R(1/(x2+y2+z2)) dv =[-22 to 22][-(8-x2) to (8-x2)][(x2+y2) to (16-x2-y2)] 1/(x2+y2+z2)dz dy dx

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in cylindrical coordinates

x=rcos, y=rsin

x2+y2=r2

(x2+y2)=8=(22)2

0<=<=2,0<=r<=22 ,r<=z<=(16-r2)

dv =r dz dr d

R(1/(x2+y2+z2)) dv =[0 to 2] [0 to 22] [r to (16-r2)] (1/(r2+z2))r dz dr d

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in spherical coordinates

x=sincos,y=sinsin,z=cos

x2+y2+z2=2

x2+y2+z2=42

==>0<=<=4

z=(x2+y2)

cos=sin

=/4

dv=2sin d d d

0<=<=2,0<=<=/4,0<=<=4

   R(1/(x2+y2+z2)) dv =[0 to 2] [0 to /4] [0 to 4] (1/2)2sin d d d

calculating using spherical coordinates is easy .because integrals are easy to calculate. i prefer that

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