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If the domain D is normal with respect to the x-axis, and f : D → R is a continuous function; then α(x) and β(x) (both of which are defined on the interval [a, b]) are the two functions that determine D. Then, by Fubini's theorem:^[5]

${\iint }_{D}f(x,y)dxdy={\int }_a^{b}dx{\int }_{\alpha \left(x\right)}^{\beta \left(x\right)}f(x,y)dy$.

If D is normal with respect to the y-axis and f : D → R is a continuous function; then α(y) and β(y) (both of which are defined on the interval [a, b]) are the two functions that determine D. Again, by Fubini's theorem:

${\iint }_{D}f(x,y)dxdy={\int }_a^{b}dy{\int }_{\alpha \left(y\right)}^{\beta \left(y\right)}f(x,y)dx$.

If T is a domain that is normal with respect to the xy-plane and determined by the functions α(x, y) and β(x, y), then

${\iiint }_{T}f(x,y,z)dxdydz={\iint }_D{\int }_{\alpha (x,y)}^{\beta (x,y)}f(x,y,z)dzdxdy$.

This definition is the same for the other five normality cases on R³. It can be generalized in a straightforward way to domains in Rⁿ.

In R² if the domain has a circular symmetry and the function has some particular characteristics one can apply the transformation to polar coordinates (see the example in the picture) which means that the generic points P(x, y) in Cartesian coordinates switch to their respective points in polar coordinates. That allows one to change the shape of the domain and simplify the operations.

The fundamental relation to make the transformation is the following:

$f(x,y)\to f(\rho \cos \phi ,\rho \sin \phi )$.

Example 2a. The function is f(x, y) = x + y and applying the transformation one obtains

$f(x,y)=f(\rho \cos \phi ,\rho \sin \phi )=\rho \cos \phi +\rho \sin \phi =\rho (\cos \phi +\sin \phi )$.

Example 2b. The function is f(x, y) = x² + y², in this case one has:

$f(x,y)={\rho }^2({\cos }^2\phi +{\sin }^2\phi )={\rho }^2$

using the Pythagorean trigonometric identity (can be useful to simplify this operation).

The transformation of the domain is made by defining the radius' crown length and the amplitude of the described angle to define the ρ, φ intervals starting from x, y.

Example 2c. The domain is D = {x² + y² ≤ 4}, that is a circumference of radius 2; it's evident that the covered angle is the circle angle, so φ varies from 0 to 2π, while the crown radius varies from 0 to 2 (the crown with the inside radius null is just a circle).

Example 2d. The domain is D = {x² + y² ≤ 9, x² + y² ≥ 4, y ≥ 0}, that is the circular crown in the positive y half-plane (please see the picture in the example); φ describes a plane angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following rectangle:

$T=\{2\le \rho \le 3,0\le \phi \le \pi \}$.

The Jacobian determinant of that transformation is the following:

$\frac{\partial (x,y)}{\partial (\rho ,\phi )}=\left|\begin{array}{cc}\cos \phi & -\rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{array}\right|=\rho$,

which has been obtained by inserting the partial derivatives of x = ρ cos(φ), y = ρ sin(φ) in the first column respect to ρ and in the second respect to φ, so the dx dy differentials in this transformation become ρ dρ dφ.

Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates:

${\iint }_{D}f(x,y)dxdy={\iint }_{T}f(\rho \cos \phi ,\rho \sin \phi )\rho d\rho d\phi$.

φ is valid in the [0, 2π] interval while ρ, which is a measure of a length, can only have positive values.

Example 2e. The function is f(x, y) = x and the domain is the same as in Example 2d. From the previous analysis of D we know the intervals of ρ (from 2 to 3) and of φ (from 0 to π). Now we change the function:

$f(x,y)=x⟶f(\rho ,\phi )=\rho \cos \phi$.

Finally let's apply the integration formula:

${\iint }_{D}xdxdy={\iint }_T\rho \cos \phi \rho d\rho d\phi$.

Once the intervals are known, you have

${\int }_0^{\pi }{\int }_2^3{\rho }^2\cos \phi d\rho d\phi ={\int }_0^{\pi }\cos \phi d\phi {\left[\frac{{\rho }^3}{3}\right]}_2^3=[\sin \phi {]}_0^{\pi }(9-\frac{8}{3})=0$.

In R³ the integration on domains with a circular base can be made by the passage to cylindrical coordinates; the transformation of the function is made by the following relation:

$f(x,y,z)\to f(\rho \cos \phi ,\rho \sin \phi ,z)$

The domain transformation can be graphically attained, because only the shape of the base varies, while the height follows the shape of the starting region.

Example 3a. The region is D = {x² + y² ≤ 9, x² + y² ≥ 4, 0 ≤ z ≤ 5} (that is the "tube" whose base is the circular crown of Example 2d and whose height is 5); if the transformation is applied, this region is obtained:

$T=\{2\le \rho \le 3,0\le \phi \le 2\pi ,0\le z\le 5\}$

(that is, the parallelepiped whose base is similar to the rectangle in Example 2d and whose height is 5).

Because the z component is unvaried during the transformation, the dx dy dz differentials vary as in the passage to polar coordinates: therefore, they become ρ dρ dφ dz.

Finally, it is possible to apply the final formula to cylindrical coordinates:

${\iiint }_{D}f(x,y,z)dxdydz={\iiint }_{T}f(\rho \cos \phi ,\rho \sin \phi ,z)\rho d\rho d\phi dz$.

This method is convenient in case of cylindrical or conical domains or in regions where it is easy to individuate the z interval and even transform the circular base and the function.

Example 3b. The function is f(x, y, z) = x² + y² + z and as integration domain this cylinder: D = {x² + y² ≤ 9, −5 ≤ z ≤ 5}. The transformation of D in cylindrical coordinates is the following:

$T=\{0\le \rho \le 3,0\le \phi \le 2\pi ,-5\le z\le 5\}$.

while the function becomes

$f(\rho \cos \phi ,\rho \sin \phi ,z)={\rho }^2+z$.

Finally one can apply the integration formula:

${\iiint }_D(x^2+y^2+z)dxdydz={\iiint }_T({\rho }^2+z)\rho d\rho d\phi dz$;

developing the formula you have

${\int }_{-5}^{5}dz{\int }_0^{2\pi }d\phi {\int }_0^3({\rho }^3+\rho z)d\rho =2\pi {\int }_{-5}^5{[\frac{{\rho }^4}{4}+\frac{{\rho }^{2}z}{2}]}_0^{3}dz=2\pi {\int }_{-5}^5(\frac{81}{4}+\frac{9}{2}z)dz=\cdots =405\pi$.

In R³ 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:

$f(x,y,z)⟶f(\rho \cos \theta \sin \phi ,\rho \sin \theta \sin \phi ,\rho \cos \phi )$.

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

The better integration domain for this passage is the sphere.

Example 4a. The domain is D = x² + y² + z² ≤ 16 (sphere with radius 4 and center at the origin); applying the transformation you get the region

$T=\{0\le \rho \le 4,0\le \phi \le \pi ,0\le \theta \le 2\pi \}$.

The Jacobian determinant of this transformation is the following:

$\frac{\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )}=\left|\begin{array}{ccc}\cos \theta \sin \phi & \rho \cos \theta \cos \phi & -\rho \sin \theta \sin \phi \\ \sin \theta \sin \phi & \rho \sin \theta \cos \phi & \rho \cos \theta \sin \phi \\ \cos \phi & -\rho \sin \phi & 0\end{array}\right|={\rho }^2\sin \phi$.

The dx dy dz differentials therefore are transformed to ρ² sin(φ) dρ dθ dφ.

This yields the final integration formula:

${\iiint }_{D}f(x,y,z)dxdydz={\iiint }_{T}f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi ){\rho }^2\sin \phi d\rho d\theta d\phi$.

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 R³ (see Example 4b); in other cases it can be better to use cylindrical coordinates (see Example 4c).

${\iiint }_{T}f(a,b,c){\rho }^2\sin \phi d\rho d\theta d\phi$.

The extra ρ² 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) = x² + y² + z² is the function to integrate. Its transformation is very easy:

$f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )={\rho }^2$,

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

$T=\{0\le \rho \le 4,0\le \phi \le \pi ,0\le \theta \le 2\pi \}$.

We therefore apply the integration formula:

${\iiint }_D(x^2+y^2+z^2)dxdydz={\iiint }_T{\rho }^2{\rho }^2\sin \theta d\rho d\theta d\phi$,

and, developing, we get

${\iiint }_T{\rho }^4\sin \theta d\rho d\theta d\phi ={\int }_0^{\pi }\sin \phi d\phi {\int }_0^4{\rho }^{4}d\rho {\int }_0^{2\pi }d\theta =2\pi {\int }_0^{\pi }\sin \phi {\left[\frac{{\rho }^5}{5}\right]}_0^{4}d\phi =2\pi {\left[\frac{{\rho }^5}{5}\right]}_0^4[-\cos \phi {]}_0^{\pi }=\frac{4096\pi }{5}$.

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

$D=\{x^2+y^2+z^2\le 9a^2\}$,

and f(x, y, z) = x² + y² 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:

$T=\{0\le \rho \le 3a,0\le \phi \le 2\pi ,0\le \theta \le \pi \}$.

However, applying the transformation, we get

$f(x,y,z)=x^2+y^2⟶{\rho }^2{\sin }^2\theta {\cos }^2\phi +{\rho }^2{\sin }^2\theta {\sin }^2\phi ={\rho }^2{\sin }^2\theta$.

Applying the formula for integration we obtain:

${\iiint }_T{\rho }^2{\sin }^2\theta {\rho }^2\sin \theta d\rho d\theta d\phi ={\iiint }_T{\rho }^4{\sin }^3\theta d\rho d\theta d\phi$,

which can be solved by turning it into an iterated integral.${\iiint }_T{\rho }^4{\sin }^3\theta d\rho d\theta d\phi =\underset{I}{\underbrace{{\int }_0^{3a}{\rho }^{4}d\rho }}\underset{II}{\underbrace{{\int }_0^{\pi }{\sin }^3\theta d\theta }}\underset{III}{\underbrace{{\int }_0^{2\pi }d\phi }}$.

$I={{\int }_0^{3a}{\rho }^{4}d\rho =\frac{{\rho }^5}{5}|}_0^{3a}=\frac{243}{5}{a}^5$,

$II={\int }_0^{\pi }{\sin }^3\theta d\theta =-{\int }_0^{\pi }{\sin }^2\theta d(\cos \theta )={\int }_0^{\pi }({\cos }^2\theta -1)d(\cos \theta )={\frac{{\cos }^3\theta }{3}|}_0^{\pi }-{\cos \theta |}_0^{\pi }=\frac{4}{3}$,

$III={\int }_0^{2\pi }d\phi =2\pi$.

Collecting all parts,

${\iiint }_T{\rho }^4{\sin }^3\theta d\rho d\theta d\phi =I\cdot II\cdot III=\frac{243}{5}{a}^5\cdot \frac{4}{3}\cdot 2\pi =\frac{648}{5}\pi a^5$.

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

$T=\{0\le \rho \le 3a,0\le \phi \le 2\pi ,-\sqrt{9a^2-{\rho }^2}\le z\le \sqrt{9a^2-{\rho }^2}\}$;

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 x² + y² into ρ²). The new function is simply ρ². Applying the integration formula

${\iiint }_T{\rho }^2\rho d\rho d\phi dz$.

Then we get:

$\begin{array}{cc}{\int }_0^{2\pi }d\phi {\int }_0^{3a}{\rho }^{3}d\rho {\int }_{-\sqrt{9a^2-{\rho }^2}}^{\sqrt{9a^2-{\rho }^2}}dz& =2\pi {\int }_0^{3a}2{\rho }^3\sqrt{9a^2-{\rho }^2}d\rho \\ & =-2\pi {\int }_{9a^2}^0(9a^2-t)\sqrt{t}dt& & t=9a^2-{\rho }^2\\ & =2\pi {\int }_0^{9a^2}(9a^2\sqrt{t}-t\sqrt{t})dt\\ & =2\pi ({\int }_0^{9a^2}9a^2\sqrt{t}dt-{\int }_0^{9a^2}t\sqrt{t}dt)\\ & =2\pi {[9a^2\frac{2}{3}{t}^{\frac{3}{2}}-\frac{2}{5}{t}^{\frac{5}{2}}]}_0^{9a^2}\\ & =2\cdot 27\pi a^5(6-\frac{18}{5})\\ & =\frac{648\pi }{5}{a}^5\end{array}$

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.