Calculus of One and Several Variables

25.3.2 Volume And Integrals in Spherical Coordinates

What is the increment of volume in spherical coordinates? There are two ways to see what this is, through art and through a systematic procedure. First consider art. Here is a picture.

In the picture there are two concentric spheres formed by making ρ two different constants and surfaces which correspond to θ assuming two different constants and ϕ assuming two different constants. These intersecting surfaces form the little box in the picture. Here is a more detailed blow up of the little box.

What is the volume of this little box? Length ≈ ρdϕ, width ≈ ρsin

dθ, height ≈ dρ and so the volume increment for spherical coordinates is

dV = ρ2sin (ϕ )dρdθdϕ

Now what is really going on? Consider the dot in the picture of the little box. Fixing θ and ϕ at their values at this point and differentiating with respect to ρ leads to a little vector of the form

( sin(ϕ)cos(θ)) ( sin(ϕ)sin(θ)) dρ cos(ϕ)

which points out from the surface of the sphere. Next keeping ρ and θ constant and differentiating only with respect to ϕ leads to an infinitesimal vector in the direction of a line of longitude,

( ρcos(ϕ)cos(θ) ) ( ρcos(ϕ)sin (θ) ) dϕ − ρ sin(ϕ)

and finally keeping ρ and ϕ constant and differentiating with respect to θ leads to the third infinitesimal vector which points in the direction of a line of latitude.

( ) − ρsin (ϕ)sin(θ) ( ρsin(ϕ)cos(θ) ) dθ 0

To find the increment of volume, we just need to take the absolute value of the determinant which has these vectors as columns, (Remember this is the absolute value of the box product.) exactly as was the case for polar coordinates. This will also yield

dV = ρ2sin(ϕ)dρdθdϕ.

However, in contrast to the drawing of pictures, this procedure is completely general and will handle all curvilinear coordinate systems and in any dimension. This is discussed more later.

In this case, U = (0,R] ×

× [0,2π) and use spherical coordinates. Then this yields a set in ℝ³ which clearly differs from the ball of radius R only by a set having volume equal to zero. It leaves out the point at the origin is all. Therefore, the volume of the ball is

The reason this was effortless, is that the ball, B_R is realized as a box in terms of the spherical coordinates. Remember what was pointed out earlier about setting up iterated integrals over boxes.

As for the integral, it is no harder to set up. You know from the transformation equations that z = ρcosϕ. Then you want

∫ ∫ ∫ ∫ R π 2π 2 2 -4 5 BR zdV = 0 0 0 (ρcos(ϕ)) ρ sinϕdθdϕd ρ = 15πR

This will be pretty easy also although somewhat more messy because the function you are integrating is not just 1 as it is when you find the volume.

Use spherical coordinates. This volume is then

∫ ∫ ∫ π 2π Rρ2 sin(ϕ)dρdθdϕ = 2πR3 + 1√3-πR3 π∕6 0 0 3 3

Now change the example a little by cutting out a cone at the bottom which has an angle of π∕2 as shown. What is the volume of what is left?

This time you would have the volume equals

∫ 3π∕4∫ 2π∫ R 1√- 1√ - ρ2sin (ϕ)dρdθdϕ = - 2πR3 + - 3πR3 π∕6 0 0 3 3

Using spherical coordinates, this gives for the volume

∫ ∫ ∫ π 2π R 2 7 3 0 π∕4 0 ρ sin (ϕ)dρdθdϕ = 6πR

This time you need

∫ 3π∕4 ∫ 2π ∫ R 7 √- 7√ - ρ2sin (ϕ)dρdθdϕ = 24 2πR3 + 24 3πR3 π∕6 π∕4 0

As to the integral, it equals

∫ ∫ ∫ 3π∕4 2π R 2 -1-√- 4( √ - ) π∕6 π∕4 0 (ρ sin(ϕ)cos(θ))ρ sin(ϕ)dρdθdϕ = − 192 2R 7π + 3 3+ 6

This is because, in terms of spherical coordinates, x = ρsin

cos.

This is entirely the wrong coordinate system to use for this problem but it is a good exercise. Here is a side view.

You need to figure out what ρ is as a function of ϕ which goes from 0 to π∕4. You should get

∫ ∫ ∫ 2π π∕4 4sec(ϕ) 2 64 0 0 0 ρ sin(ϕ)dρdϕdθ = 3 π

As to ∫ _RzdV, it equals

∫ 2π∫ π∕4∫ 4sec(ϕ)◜--z◞◟-◝ ρ cos(ϕ)ρ2sin (ϕ)dρdϕdθ = 64π 0 0 0

In  cylindrical coordinates,

( ) ( ) x rcosθ ( y ) = ( rsinθ ) z z

Therefore, the Jacobian determinant is

( ) ( cosθ − r sinθ 0 ) det sinθ rcosθ 0 = r. 0 0 1

It follows the volume element in cylindrical coordinates is rdθdrdz.

This is a better coordinate system for this example than spherical coordinates. This time you should get

∫ 2π ∫ 4∫ 4 rdzdrdθ = 64 π 0 0 r 3

Without loss of generality, assume z₀ > 0. Let dV be a little chunk of material located at the point

of H the hollow sphere. Then according to Newton’s law of gravity, the force this small chunk of material exerts on the given point mass equals

-xi+-yj+-(z −-z0)k-(-------1-------)G αdV = |xi+ yj+ (z − z0)k| x2 +y2 + (z − z0)2

----------1---------- (xi+ yj+ (z − z0)k)( 2 2 2)3∕2GαdV x + y + (z − z0)

Therefore, the total force is

∫ (xi+ yj+ (z − z0)k) (-------1------)---G αdV. H x2 + y2 + (z − z0)2 3∕2

By the symmetry of the sphere, the i and j components will cancel out when the integral is taken. This is because there is the same amount of stuff for negative x and y as there is for positive x and y. Hence what remains is

∫ αGk -------(z −-z0)------dV H [ 2 2 2]3∕2 x + y + (z − z0)

as claimed. Now for the interesting part, the integral is evaluated. In spherical coordinates this integral is.

∫ ∫ ∫ 2π b π -(ρ-cosϕ-−-z0)ρ2sinϕ--- 0 a 0 (ρ2 + z2− 2ρz cosϕ)3∕2dϕdρdθ. (25.1) 0 0

(25.1)

Rewrite the inside integral and use integration by parts to obtain this inside integral equals

∫ π -1- (ρ2cosϕ − ρz0) -----(2z0ρsin-ϕ)-----dϕ = 2z0 0 (ρ2 + z20 − 2ρz0cosϕ)3∕2

( -1- ----− ρ2-−-ρz0--- -----ρ2 −-ρz0--- 2z0 − 2∘ (ρ2 +-z2+-2ρz-) + 2∘ (ρ2 +-z2-− 2ρz-) 0 0 0 0

∫ π ) − 2ρ2∘-------sin-ϕ--------dϕ . (25.2) 0 (ρ2 + z20 − 2ρz0cosϕ)

(25.2)

There are some cases to consider here.

First suppose z₀ < a so the point is on the inside of the hollow sphere and it is always the case that ρ > z₀. Then in this case, the two first terms reduce to

2ρ (ρ + z0) 2ρ (ρ − z0) 2ρ (ρ+ z0) 2ρ(ρ − z0) ∘--------2 + ∘-------2 = -(ρ-+-z)--+ --ρ−-z--- = 4ρ (ρ+ z0) (ρ− z0) 0 0

and so the expression in 25.2 equals

1 ( ∫ π sin ϕ ) --- 4ρ − 2ρ2∘---2---2-----------dϕ 2z0 0 (ρ + z0 − 2ρz0cosϕ)

1 ( 1 ∫ π 2ρz sinϕ ) = --- 4ρ− -- ρ∘---2---20----------dϕ 2z0 z0 0 (ρ + z0 − 2ρz0cosϕ)

( ) = -1- 4ρ− 2ρ(ρ2 + z2− 2ρz cosϕ)1∕2|π 2z0 z0 0 0 0

1 ( 2ρ ) = 2z- 4ρ − z-[(ρ + z0)− (ρ − z0)] = 0. 0 0

Therefore, in this case the inner integral of 25.1 equals zero and so the original integral will also be zero.

The other case is when z₀ > b and so it is always the case that z₀ > ρ. In this case the first two terms of 25.2 are

2∘ρ(ρ+-z0)-+ 2∘ρ(ρ-− z0)-= 2ρ(ρ-+z0) + 2ρ(ρ−-z0)= 0. (ρ + z0)2 (ρ− z0)2 (ρ+ z0) z0 − ρ

Therefore in this case, 25.2 equals

which equals

− ρ(( )1∕2 ) − ρ 2ρ2 z2- ρ2 + z20 − 2ρz0cosϕ |π0 = -z2[(ρ+ z0)− (z0 − ρ)] = −-z2. 0 0 0

Thus the inner integral of 25.1 reduces to the above simple expression. Therefore, 25.1 equals

∫ ∫ ( ) 2π b 2- 2 4 b3 −-a3 0 a − z20ρ dρdθ = − 3π z20

and so