Triple Integrals: Cartesian coordinates into spherical polar coordinates

CARTESIAN TO SPHERICAL POLAR COORDINATES

Let us first define spherical polar coordinates of a point in space and derive the relations between Cartesian and spherical polar coordinates for above figure

Let P be the point whose Cartesian coordinate are (x, y, z).

Let PM be drawn ⟂ r to the xOy plane. Let MN parallel to y = axis. Let = r, the angle made by OP with the positive z‒axis = θ and angle made by OM with x‒axix = Ø.

The triplet (r, θ, Ø) are called the spherical polar coordinate of P.

Since OMP = 90°, MP = z = r cosθ, OM = r sinθ, ON = x = r sinθcosØ and NM = y = r sinθsinØ

Thus the transformations from three dimensional Cartesian to spherical polar coordinates are

x = r sinθ cosØ

y = r sinθ sin Ø.

z = r cosθ

Also x² + y²+z² = r²,

In this case,

J = ∂(x,y,z) / ∂(r,θ,z) = r² sin 0

Hence dx dy dz = r² sinθ dr dθ dz

and ∫∫∫_V f(x, y, z) dx dy dz = ∫∫∫_V g(r,θ,z) r² sinθ dr dθ dz

Note: Whenever ∫∫∫ f(x, y, z) dx dy dz is to be evaluated throughout the volume of a sphere, hemisphere or octant of a sphere, it will be advantageous to use spherical polar coordinates.

Example 82. Change the spherical polar coordinates and hence evaluate ∫∫∫_V dxdydz / x²+y²+z² , where V is the volume of the sphere x² + y² + z² = a².

Solution:

The given integral is

I = ∫∫∫_V dxdydz / x²+y²+z²

To convert from Cartesian to spherical polar coordinates system, we have the following transformation

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

dxdydz = r² sinθ drdθdϕ

Also x² + y²+z² = r².

The region is entire volume of the sphere x² + y² + z² = a².

Therefore the limits are

r=0 ; r=a

θ = 0 ; θ = π

ϕ = 0 ; ϕ = 2π

I = 4πα

Example 83. Evaluate ∫∫∫ √[1‒x² ‒ y² ‒ z²] dx dy dz, taken throughout the volume of the sphere x² + y²+z² = 1, by transforming to spherical polar coordinates.

Solution:

The given integral is

I = ∫∫∫ √[1‒x² ‒ y² ‒ z²] dx dy dz

To convert from Cartesian to spherical polar coordinates system, we have the following transformation

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

dxdydz= r² sinθ drdθdϕ

Also x² + y² + z² = r².

The region is entire volume of the sphere x² + y² + z² = 1.

Therefore the limits are

r=0 ; r=1

θ=0 ; θ = π

ϕ=0 ; ϕ = 2π

Put r = sint in the innermost integral

dr = cost dt

r = 0 ⇒ sint = 0 ⇒ t=0

r = 1 ⇒ sint = 0 ⇒ t = π/2

Example 84. Evaluate , by transforming to spherical polar coordinates.

Solution:

The boundaries of the region of integration are z = 0, z = √[a²‒x² ‒ y²] or x² + y²+z² = a², y = 0, y = √[a² ‒ x²] or x² + y² = a², x = 0 and x = a. From the boundaries, we note that the region of integration is the volume of the positive octant of the sphere x² + y² + z² = a².

To convert from Cartesian to spherical polar coordinates system, we have the following transformation

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cos θ

dxdydz = r² sinθ drdθdϕ

Also x² + y² + z² = r².

The region is volume of the positive octant of the sphere x² + y² + z² = a².

Therefore the limits are

r=0 ; r = a

θ = 0 ; θ = π/2

ϕ = 0 ; ϕ = π/2

Example 85. Evaluate  using spherical polar coordinates.

Solution:

The given integral is

The limits are

 z = √[x² + y²] ;  z = 1

 y=0 ;  y = √[1‒x²]

 x=0 ;  x = 1

To convert from Cartesian to spherical polar coordinates system, we have the following transformation

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

dxdydz = r² sinθ drdθdϕ

Also x² + y² + z² = r².

We have z = 1

r cosθ = 1

r = 1 / cosθ = secθ

We have z = √[x² + y²] ⇒ r cosθ = √[ (r sinθ cosϕ )² + (r sinθ sinϕ)²]

⇒ r cosθ = √[ (r sinθ)² [cos²ϕ + sin²ϕ] ]

⇒ r cosθ = r sinθ ⇒ tanθ = 1⇒ θ = π/4

Since both lower limits of x and y are zero, ϕ = 0 to ϕ = π/2

Therefore the limits are

r = 0 ; r = secθ

θ = 0 ; θ = π/4

ϕ = 0 ; ϕ = π/2

Example 86. Evaluate the integral ∫∫∫ (x² + y² + z²) dx dy dz taken over the volume enclosed by the sphere x² + y² + z² =1.

Solution:

Since the region is a sphere, we use spherical polar co‒ordinates.

The spherical polar co‒ordinates is

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

x² + y²+z² = r²

dxdydz = r² sinθ drdθdϕ

For the entire sphere x² + y²+z² = 1

Limits are

r=0 to r = 1

θ=0 to θ = π

ϕ = 0 to ϕ = 2π

Evaluate the integration ∫∫∫ xyz dx dy dz taken throughout the volume for which x, y, z ≥ 0 and x² + y² + z² ≤ 9.

Solution:

Since the region is a sphere, we use spherical polar co‒ordinates.

The spherical polar co‒ordinates is

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

x² + y²+z² = r²

dxdydz = r² sinθ drdθdϕ

For first octant of the sphere x² + y²+z² = 9

Limits are

r = 0 to r = 3

θ = 0 to θ = π/2

ϕ = 0 to ϕ = π/2

EXERCISE

77. Evaluate the integrals by changing to cylindrical co‒ordinates

Ans: 0

78. Evaluate the integrals by changing to cylindrical co‒ordinates.

Ans: 4√2‒5 / 15

79. Evaluate ∫∫∫_V (x² + y²)dV, V lies between the spheres x² + y² + z² = 4 and x²+ y²+z² = 9. Ans: 1688π / 15.

80. Evaluate the integral ∫∫∫ (x² + y²+z²) dx dy dz taken over the volume closed of the sphere x² + y²+ z² = a² transforming into spherical.co‒ ordinates.

Ans: 4π/5 a⁵

81. Evaluate the integration ∫∫∫ xyz dz dy dx through the spherical octant for which x² + y²+z² ≤ a². Ans: a⁶/48