Triple Integrals in Cylindrical and Spherical Coordinates

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Cylindrical coordinates

Cylindrical coordinates.

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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 = {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:

(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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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) = 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = \{ 0 \le \rho \le 3, \ 0 \le \varphi \le 2 \pi, \ -5 \le z \le 5 \}.}

while the function becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\rho \cos \varphi, \rho \sin \varphi, z) = \rho^2 + z}

Finally one can apply the integration formula:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 2Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} .

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = \{ 0 \le \rho \le 4, \ 0 \le \varphi \le \pi, \ 0 \le \theta \le 2 \pi \}.}

The Jacobian determinant of this transformation is the following:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 ρ2 sin(φ) dρ dθ dφ.

This yields the final integration formula:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 R3 (see Example 4b); in other cases it can be better to use cylindrical coordinates (see Example 4c).

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \iiint_T f(a,b,c) \rho^2 \sin \varphi \, d\rho\, d\theta\, d\varphi.}

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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=\{0 \le \rho \le 4, \ 0 \le \varphi \le \pi, \ 0 \le \theta \le 2 \pi\}.}

We therefore apply the integration formula:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 3a,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D = \left \{ x^2 + y^2 + z^2 \le 9a^2 \right \}}

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 obviously:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=\{0 \le \rho \le 3a, \ 0 \le \varphi \le 2 \pi, \ 0 \le \theta \le \pi\}.}

However, applying the transformation, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \left.\int_0^{3a}\rho^4 d\rho = \frac{\rho^5}{5}\right\vert_0^{3a} = \frac{243}{5}a^5} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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|^\pi_0 - \left.\cos\theta\right|^\pi_0 = \frac{4}{3}} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle III = \int_0^{2\pi} d \varphi = 2\pi} .


Collecting all parts,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=\left\{0 \le \rho \le 3a, \ 0 \le \varphi \le 2 \pi, \ - \sqrt{9a^2 - \rho^2} \le z \le \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 x2 + y2 into ρ2). The new function is simply ρ2. Applying the integration formula

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \iiint_T \rho^2 \rho \, d \rho \, d \varphi \, dz.}

Then we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_0^{2\pi} d\varphi \int_0^{3a} \rho^3 d\rho \int_{-\sqrt{9a^2 - \rho^2}}^{\sqrt{9 a^2 - \rho^2}}\, dz &= 2 \pi \int_0^{3a} 2 \rho^3 \sqrt{9 a^2 - \rho^2} \, d\rho \\ &= -2 \pi \int_{9 a^2}^0 (9 a^2 - t) \sqrt{t}\, dt && t = 9 a^2 - \rho^2 \\ &= 2 \pi \int_0^{9 a^2} \left ( 9 a^2 \sqrt{t} - t \sqrt{t} \right ) \, dt \\ &= 2 \pi \left( \int_0^{9 a^2} 9 a^2 \sqrt{t} \, dt - \int_0^{9 a^2} t \sqrt{t} \, dt\right) \\ &= 2 \pi \left[9 a^2 \frac23 t^{ \frac32 } - \frac{2}{5} t^{ \frac{5}{2}} \right]_0^{9 a^2} \\ &= 2 \cdot 27 \pi a^5 \left ( 6 - \frac{18}{5} \right ) \\ &= \frac{648 \pi}{5} a^5. \end{align}}

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

Resources

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