Triple Integrals in Cylindrical and Spherical Coordinates

Cylindrical coordinates

Cylindrical coordinates.

In $R 3$ 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)\rightarrow f(\rho \cos \varphi ,\rho \sin \varphi ,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 $D = {x 2 + y 2 \leq 9, x 2 + y 2 \geq 4, 0 \leq z \leq 5}$ (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:
$T=\{2\leq \rho \leq 3,\ 0\leq \varphi \leq 2\pi ,\ 0\leq z\leq 5\}$

(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:

$\iiint _{D}f(x,y,z)\,dx\,dy\,dz=\iiint _{T}f(\rho \cos \varphi ,\rho \sin \varphi ,z)\rho \,d\rho \,d\varphi \,dz.$
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 $f (x, y, z) = x 2 + y 2 + z$ and as integration domain this cylinder: $D = {x 2 + y 2 \leq 9, -5 \leq z \leq 5}$ . The transformation of $D$ in cylindrical coordinates is the following:
$T=\{0\leq \rho \leq 3,\ 0\leq \varphi \leq 2\pi ,\ -5\leq z\leq 5\}.$

while the function becomes

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

Finally one can apply the integration formula:

$\iiint _{D}\left(x^{2}+y^{2}+z\right)\,dx\,dy\,dz=\iiint _{T}\left(\rho ^{2}+z\right)\rho \,d\rho \,d\varphi \,dz;$

developing the formula you have

$\int _{-5}^{5}dz\int _{0}^{2\pi }d\varphi \int _{0}^{3}\left(\rho ^{3}+\rho z\right)\,d\rho =2\pi \int _{-5}^{5}\left[{\frac {\rho ^{4}}{4}}+{\frac {\rho ^{2}z}{2}}\right]_{0}^{3}\,dz=2\pi \int _{-5}^{5}\left({\frac {81}{4}}+{\frac {9}{2}}z\right)\,dz=\cdots =405\pi .$

Spherical coordinates

Spherical coordinates.

In $R 3$ 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)\longrightarrow f(\rho \cos \theta \sin \varphi ,\rho \sin \theta \sin \varphi ,\rho \cos \varphi )

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

The better integration domain for this passage is the sphere.

Example 4a. The domain is $D = x 2 + y 2 + z 2 \leq 16$ (sphere with radius 4 and center at the origin); applying the transformation you get the region
$T=\{0\leq \rho \leq 4,\ 0\leq \varphi \leq \pi ,\ 0\leq \theta \leq 2\pi \}.$

The Jacobian determinant of this transformation is the following:

${\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}={\begin{vmatrix}\cos \theta \sin \varphi &-\rho \sin \theta \sin \varphi &\rho \cos \theta \cos \varphi \\\sin \theta \sin \varphi &\rho \cos \theta \sin \varphi &\rho \sin \theta \cos \varphi \\\cos \varphi &0&-\rho \sin \varphi \end{vmatrix}}=\rho ^{2}\sin \varphi$

The $dx dy dz$ differentials therefore are transformed to $ρ 2 sin(φ) dρ dθ dφ$ .
This yields the final integration formula:

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

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 $R 3$ (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 \varphi \,d\rho \,d\theta \,d\varphi .

The extra $ρ 2$ 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 $f (x, y, z) = x 2 + y 2 + z 2$ is the function to integrate. Its transformation is very easy:
$f(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi )=\rho ^{2},$

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

$T=\{0\leq \rho \leq 4,\ 0\leq \varphi \leq \pi ,\ 0\leq \theta \leq 2\pi \}.$

We therefore apply the integration formula:

$\iiint _{D}\left(x^{2}+y^{2}+z^{2}\right)\,dx\,dy\,dz=\iiint _{T}\rho ^{2}\,\rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi ,$

and, developing, we get

$\iiint _{T}\rho ^{4}\sin \theta \,d\rho \,d\theta \,d\varphi =\int _{0}^{\pi }\sin \varphi \,d\varphi \int _{0}^{4}\rho ^{4}d\rho \int _{0}^{2\pi }d\theta =2\pi \int _{0}^{\pi }\sin \varphi \left[{\frac {\rho ^{5}}{5}}\right]_{0}^{4}\,d\varphi =2\pi \left[{\frac {\rho ^{5}}{5}}\right]_{0}^{4}{\Big [}-\cos \varphi {\Big ]}_{0}^{\pi }={\frac {4096\pi }{5}}.$

Example 4c. The domain $D$ is the ball with center at the origin and radius $3 a$ ,
$D=\left\{x^{2}+y^{2}+z^{2}\leq 9a^{2}\right\}$

and $f (x, y, z) = x 2 + y 2$ 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 obviously:

$T=\{0\leq \rho \leq 3a,\ 0\leq \varphi \leq 2\pi ,\ 0\leq \theta \leq \pi \}.$

However, applying the transformation, we get

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

Applying the formula for integration we obtain:

$\iiint _{T}\rho ^{2}\sin ^{2}\theta \rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi =\iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi$

which can be solved by turning it into an iterated integral.

$\iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =\underbrace {\int _{0}^{3a}\rho ^{4}d\rho } _{I}\,\underbrace {\int _{0}^{\pi }\sin ^{3}\theta \,d\theta } _{II}\,\underbrace {\int _{0}^{2\pi }d\varphi } _{III}$ .
$I=\left.\int _{0}^{3a}\rho ^{4}d\rho ={\frac {\rho ^{5}}{5}}\right\vert _{0}^{3a}={\frac {243}{5}}a^{5}$ ,
$II=\int _{0}^{\pi }\sin ^{3}\theta \,d\theta =-\int _{0}^{\pi }\sin ^{2}\theta \,d(\cos \theta )=\int _{0}^{\pi }(\cos ^{2}\theta -1)\,d(\cos \theta )=\left.{\frac {\cos ^{3}\theta }{3}}\right|_{0}^{\pi }-\left.\cos \theta \right|_{0}^{\pi }={\frac {4}{3}}$ ,
$III=\int _{0}^{2\pi }d\varphi =2\pi$ .

Collecting all parts,
$\iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =I\cdot II\cdot III={\frac {243}{5}}a^{5}\cdot {\frac {4}{3}}\cdot 2\pi ={\frac {648}{5}}\pi a^{5}$ .

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

$T=\left\{0\leq \rho \leq 3a,\ 0\leq \varphi \leq 2\pi ,\ -{\sqrt {9a^{2}-\rho ^{2}}}\leq z\leq {\sqrt {9a^{2}-\rho ^{2}}}\right\};$

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 $x 2 + y 2$ into $ρ 2$ ). The new function is simply $ρ 2$ . Applying the integration formula

$\iiint _{T}\rho ^{2}\rho \,d\rho \,d\varphi \,dz.$

Then we get

${\begin{aligned}\int _{0}^{2\pi }d\varphi \int _{0}^{3a}\rho ^{3}d\rho \int _{-{\sqrt {9a^{2}-\rho ^{2}}}}^{\sqrt {9a^{2}-\rho ^{2}}}\,dz&=2\pi \int _{0}^{3a}2\rho ^{3}{\sqrt {9a^{2}-\rho ^{2}}}\,d\rho \\&=-2\pi \int _{9a^{2}}^{0}(9a^{2}-t){\sqrt {t}}\,dt&&t=9a^{2}-\rho ^{2}\\&=2\pi \int _{0}^{9a^{2}}\left(9a^{2}{\sqrt {t}}-t{\sqrt {t}}\right)\,dt\\&=2\pi \left(\int _{0}^{9a^{2}}9a^{2}{\sqrt {t}}\,dt-\int _{0}^{9a^{2}}t{\sqrt {t}}\,dt\right)\\&=2\pi \left[9a^{2}{\frac {2}{3}}t^{\frac {3}{2}}-{\frac {2}{5}}t^{\frac {5}{2}}\right]_{0}^{9a^{2}}\\&=2\cdot 27\pi a^{5}\left(6-{\frac {18}{5}}\right)\\&={\frac {648\pi }{5}}a^{5}.\end{aligned}}$
Thanks to the passage to cylindrical coordinates it was possible to reduce the triple integral to an easier one-variable integral.