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Spherical coordinates

In R some domains have a spherical symmetry, so it's possible to specify the coordinates of every point of the integration region by two angles and one distance. It's possible to use therefore the passage to spherical coordinates; the function is transformed by this relation:

Points on the z -axis do not have a precise characterization in spherical coordinates, so θ can vary between 0 and 2 π.

The better integration domain for this passage is the sphere.

Example 4a. The domain is D = x + y + z ≤ 16 (sphere with radius 4 and center at the origin); applying the transformation you get the region

The Jacobian determinant of this transformation is the following:

The dx dy dz differentials therefore are transformed to ρ sin(φ) dρ dθ dφ.

This yields the final integration formula:

It is better to use this method in case of spherical domains and in case of functions that can be easily simplified by the first fundamental relation of trigonometry extended to R (see Example 4b); in other cases it can be better to use cylindrical coordinates (see Example 4c).

The extra ρ and sin φ come from the Jacobian.

In the following examples the roles of φ and θ have been reversed.

Example 4b. D is the same region as in Example 4a and f (x, y, z) = x + y + z is the function to integrate. Its transformation is very easy:

while we know the intervals of the transformed region T from D:

We therefore apply the integration formula:

and, developing, we get

Example 4c. The domain D is the ball with center at the origin and radius 3 a,

and f (x, y, z) = x + y is the function to integrate.

Looking at the domain, it seems convenient to adopt the passage to spherical coordinates, in fact, the intervals of the variables that delimit the new T region are:

However, applying the transformation, we get

Applying the formula for integration we obtain:

which can be solved by turning it into an iterated integral. ∭ T ρ 4 sin 3 θ d ρ d θ d φ = ∫ 0 3 a ρ 4 d ρ ⏟ I ∫ 0 π sin 3 θ d θ ⏟ I I ∫ 0 2 π d φ ⏟ I I I {\displaystyle \iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =\underbrace {\int _{0}^{3a}\rho ^{4}d\rho } _{I}\,\underbrace {\int _{0}^{\pi }\sin ^{3}\theta \,d\theta } _{II}\,\underbrace {\int _{0}^{2\pi }d\varphi } _{III}}.

I = ∫ 0 3 a ρ 4 d ρ = ρ 5 5 | 0 3 a = 243 5 a 5 {\displaystyle I=\left.\int _{0}^{3a}\rho ^{4}d\rho ={\frac {\rho ^{5}}{5}}\right\vert _{0}^{3a}={\frac {243}{5}}a^{5}},

I I = ∫ 0 π sin 3 θ d θ = − ∫ 0 π sin 2 θ d (cos θ) = ∫ 0 π (cos 2 θ − 1) d (cos θ) = cos 3 θ 3 | 0 π − cos θ | 0 π = 4 3 {\displaystyle II=\int _{0}^{\pi }\sin ^{3}\theta \,d\theta =-\int _{0}^{\pi }\sin ^{2}\theta \,d(\cos \theta)=\int _{0}^{\pi }(\cos ^{2}\theta -1)\,d(\cos \theta)=\left.{\frac {\cos ^{3}\theta }{3}}\right|_{0}^{\pi }-\left.\cos \theta \right|_{0}^{\pi }={\frac {4}{3}}},

I I I = ∫ 0 2 π d φ = 2 π {\displaystyle III=\int _{0}^{2\pi }d\varphi =2\pi }.

Collecting all parts,

∭ T ρ 4 sin 3 θ d ρ d θ d φ = I ⋅ I I ⋅ I I I = 243 5 a 5 ⋅ 4 3 ⋅ 2 π = 648 5 π a 5 {\displaystyle \iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =I\cdot II\cdot III={\frac {243}{5}}a^{5}\cdot {\frac {4}{3}}\cdot 2\pi ={\frac {648}{5}}\pi a^{5}}.

Alternatively, this problem can be solved by using the passage to cylindrical coordinates. The new T intervals are

the z interval has been obtained by dividing the ball into two hemispheres simply by solving the inequality from the formula of D (and then directly transforming x + y into ρ). The new function is simply ρ. Applying the integration formula

Then we get:

Thanks to the passage to cylindrical coordinates it was possible to reduce the triple integral to an easier one-variable integral.

See also the differential volume entry in nabla in cylindrical and spherical coordinates.