Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

Express the volume of V in Cartesian, cylindrical and spherical coordinates.?

Let V be the region in R^3 inside the sphere x^2+y^2+z^2=1 and above te plane z=0.

Express the volume of V in Cartesian, cylindrical and spherical coordinates.

I did some work on this but not sure whether they are right. Can anybody help me with this one?

Thanks!

2 Answers

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  • J
    Lv 7
    7 years ago
    Best Answer

    Cartesian

    V= INT dV = INT dz dydx

    limits

    0<z<sqrt (1- (x^2+y^2))

    Since the Region is at z=0 , then it is x^2+y^2 =1 , so

    -sqrt (1-x^2)<y<+sqrt (1-x^2)

    -1<x<1

    You can use dxdy also -sqrt (1-y^2) <x< +sqrt (1-y^2) and -1<y<+1

    Cylindricals

    x=rcosT

    y=rsinT

    z=z

    The sphere is ( plug x and y ) r^2+z^2 =1 , z= sqrt ( 1-r^2)

    Note that we are using r , not R=1 , because we are moving inside the Region x^2+y^2=1

    So , V= INT_V dz dA

    Over the Region , dA = rdrdT ( Polar coordinates )

    V= INT _R z rdrdT

    0<z<sqrt ( 1-r^2)

    0<r<R=1

    0<T<2pi

    Solving , just in case , V=INT INT rsqrt (1-r^2) drdT

    u=1-r^2

    du= -2r dr

    V= INT INT u^(1/2) (-1/2) du dT

    V= INT (2/3) u^(3/2) (-1/2) dT

    V= INT (-1/3) (1-r^2) ^(3/2) dT

    V= INT (-1/3) ( 0- 1) dT herical

    V= INT (1/3) dT

    V= 2pi/3

    Spherical

    x=rsinZcosT

    y=rsinZsinT

    z=rcosZ

    The sphere ( Plug x, y,z ) is r=1

    dV= r^2sinZ dZdrdT

    V= INT_V dV = INT INT INT r^2sinZ drdZdT

    Limits

    0<r<1

    0<Z<pi/2

    0<T<2pi

    V= INT INT r^3/3 sinZ dZdT

    V= (1/3) INT INT sinZ dZdT

    V= (1/3) (-cosZ) INT dT

    V= (1/3) (-) (0-1) INT dT

    V= (1/3) *2pi

    V= 2pi/3

    God bless you

  • 7 years ago

    Spherical: very simple r<1 and θ = 0 to π/2 and φ = 0 to 2π

    Cylindrical:-1<r<1; 0<z<1 and θ = 0 to 2π

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