Calculus

Triple Integrals

Triple Integrals in Spherical Coordinates

Page 1
Problem 1
Page 2
Problems 2-5

The spherical coordinates of a point \(M\left( {x,y,z} \right)\) are defined to be the three numbers: \(\rho, \varphi, \theta,\) where

  • \(\rho\) is the length of the radius vector to the point M;
  • \(\varphi\) is the angle between the projection of the radius vector \(\overrightarrow {OM} \) on the \(xy\)-plane and the \(x\)-axis;
  • \(\theta\) is the angle of deviation of the radius vector \(\overrightarrow {OM}\) from the positive direction of the \(z\)-axis (Figure \(1\)).

It’s important to take into account that the definition of \(\rho\) differs in spherical and cylindrical coordinates.

Spherical system of coordinates

Figure 1.

The spherical coordinates of a point are related to its Cartesian coordinates as follows:
\[
{x = \rho \cos \varphi \sin \theta ,\;\;\;}\kern0pt
{y = \rho \sin \varphi \sin \theta ,\;\;\;}\kern0pt
{z = \rho \cos \theta ,}
\]
\[
{\text{where}\;\;\rho \ge 0,\;\;\;}\kern0pt
{0 \le \varphi \le 2\pi ,\;\;\;}\kern0pt
{0 \le \theta \le \pi .}
\]
The Jacobian of transformation from Cartesian to cylindrical coordinates is written as
\[
{I\left( {\rho ,\varphi ,\theta } \right) }
= {\left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial \theta }}}\\
{\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial \theta }}}\\
{\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial \theta }}}
\end{array}} \right|,}
\]
where the partial derivatives are given by
\[
{\frac{{\partial x}}{{\partial \rho }} = \cos \varphi \sin \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial x}}{{\partial \varphi }} = – \rho \sin \varphi \sin \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial x}}{{\partial \theta }} = \rho \cos \varphi \cos \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial y}}{{\partial \rho }} = \sin \varphi \sin \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial y}}{{\partial \varphi }} = – \rho \cos \varphi \sin \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial y}}{{\partial \theta }} = \rho \sin \varphi \cos \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial z}}{{\partial \rho }} = \cos \theta,\;\;\;}\kern-0.3pt
{\frac{{\partial z}}{{\partial \varphi }} = 0,\;\;\;}\kern-0.3pt
{\frac{{\partial z}}{{\partial \theta }} = -\rho \sin \theta .}
\]
By expanding the determinant along the second column, we get
\[
{I\left( {\rho ,\varphi ,\theta } \right) }
= {{\rho \sin \varphi \sin \theta \cdot}\kern0pt{ \left| {\begin{array}{*{20}{c}}
{\sin \varphi \sin \theta }&{\rho \sin \varphi \cos \theta }\\
{\cos \theta }&{ – \rho \sin \theta }
\end{array}} \right| }}
+ {{\rho \cos \varphi \sin \theta \cdot}\kern0pt{ \left| {\begin{array}{*{20}{c}}
{\cos \varphi \sin \theta }&{\rho \cos \varphi \cos \theta }\\
{\cos \theta }&{ – \rho \sin \theta }
\end{array}} \right| }}
= {{\rho \sin \varphi \sin \theta }\kern0pt{ \left( { – \rho \sin \varphi \,{{\sin }^2}\theta – \rho \sin \varphi \,{{\cos }^2}\theta } \right) }}
+ {{\rho \cos \varphi \sin \theta }\kern0pt{ \left( { – \rho \cos \varphi \,{{\sin }^2}\theta – \rho \cos \varphi \,{{\cos }^2}\theta } \right) }}
= {\rho \sin \varphi \sin \theta \cdot \left( { – \rho \sin \varphi } \right) \cdot 1 }
+ {\rho \cos \varphi \sin \theta \cdot \left( { – \rho \cos \varphi } \right) \cdot 1 }
= {- {\rho ^2}\sin \theta \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right) }
= { – {\rho ^2}\sin \theta .}
\]
Accordingly, the absolute value of the Jacobian is
\[{\left| {I\left( {\rho ,\varphi ,\theta } \right)} \right| }={ {\rho ^2}\sin \theta .}\]
Hence, the formula of change of variables for this transformation is
\[
{\iiint\limits_U {f\left( {x,y,z} \right)dxdydz \text{ = }} }\kern-0.3pt
{\iiint\limits_{U’} {f\left( {{\rho \cos \varphi \sin \theta ,\,}{\rho \sin \varphi \sin \theta,\,} }\right.}\kern0pt{\left.{ {\rho \cos \theta }} \right) }\kern0pt{ {\rho ^2}\sin \theta d\rho d\varphi d\theta } .}
\]
It is easier to calculate triple integrals in spherical coordinates when the region of integration \(U\) is a ball (or some portion of it) and/or when the integrand is a kind of \(f\left( {{x^2} + {y^2} + {z^2}} \right).\)

It is sometimes more convenient to use so-called generalized spherical coordinates, related to the Cartesian coordinates by the formulas
\[
{x = a\rho \cos \varphi \sin \theta ,\;\;\;}\kern0pt
{y = b\rho \sin \varphi \sin \theta ,\;\;\;}\kern0pt
{z = c\rho \cos \theta .}
\]
In this case the Jacobian is
\[{I\left( {\rho ,\varphi ,\theta } \right) }={ – abc{\rho ^2}\sin \theta .}\]

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the integral \(\iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz},\) where the region of integration \(U\) is the ball given by the equation \({{x^2} + {y^2} + {z^2}} \) \(= 25.\)

 Example 2

Calculate the integral
\[\iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} ,\]
where the region \(U\) is the unit ball \({{x^2} + {y^2} + {z^2}} \le 1.\)

 Example 3

Evaluate the integral \(\iiint\limits_U {xyzdxdydz} ,\) where the region \(U\) is a portion of the ball \({x^2} + {y^2} + {z^2} \le {R^2},\) lying in the first octant \(x \ge 0, y \ge 0, z \ge 0.\)

 Example 4

Find the triple integral
\[\iiint\limits_U {\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} }\right.\,}\kern0pt{+\,\left.{ \frac{{{z^2}}}{{{c^2}}}} \right)dxdydz},\]
where the region \(U\) is bounded by the ellipsoid
\[{\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}}} = 1.\]

 Example 5

Evaluate the integral
\[{\int\limits_0^1 {dx} \int\limits_0^{\sqrt {1 – {x^2}} } {dy} }\kern0pt{\int\limits_0^{\sqrt {1 – {x^2} – {y^2}} } {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}dz} ,}\]
using spherical coordinates.

Example 1.

Evaluate the integral \(\iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz},\) where the region of integration \(U\) is the ball given by the equation \({{x^2} + {y^2} + {z^2}} \) \(= 25.\)

Solution.

As the region \(U\) is a ball and the integrand is expressed by a function depending on \(f\left( {{x^2} + {y^2} + {z^2}} \right),\) we can convert the triple integral to spherical coordinates. Make the substitution:
\[
{x = \rho \cos \varphi \sin \theta ,\;\;\;}\kern0pt
{y = \rho \sin \varphi \sin \theta ,\;\;\;}\kern0pt
{z = \rho \cos \theta ,}
\]
The new variables range within the limits:
\[
{0 \le \rho \le 5,\;\;\;}\kern0pt
{0 \le \varphi \le 2\pi ,\;\;\;}\kern0pt
{0 \le \theta \le \pi .}
\]
Taking into account the Jacobian \({\rho ^2}\sin \theta,\) we can write the integral as follows:
\[
{I = \iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz} }
= {\iiint\limits_{U’} {\rho \cdot {\rho ^2}\sin \theta d\rho d\varphi d\theta } }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \int\limits_0^\pi {\sin \theta d\theta } }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left[ {\left. {\left( { – \cos \theta } \right)} \right|_0^\pi } \right] }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left( { – \cos \pi + \cos 0} \right) }
= {2\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } }
= {2\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}} \right)} \right|_0^5} \right] }
= {2\int\limits_0^{2\pi } {d\varphi } \cdot \frac{{{5^4}}}{4} }
= {\frac{{625}}{2}\int\limits_0^{2\pi } {d\varphi } }
= {\frac{{625}}{2} \cdot 2\pi }={ 625\pi .}
\]

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Problem 1
Page 2
Problems 2-5