Calculus

Triple Integrals

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Triple Integrals in Spherical Coordinates

  • 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 a problem to see the 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:

    \[
    {I = \iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz} }
    = {\iiint\limits_{U’} {\rho \cdot {\rho ^2}\sin \theta d\rho d\varphi d\theta } }
    = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \int\limits_0^\pi {\sin \theta d\theta } }
    = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left[ {\left. {\left( { – \cos \theta } \right)} \right|_0^\pi } \right] }
    = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left( { – \cos \pi + \cos 0} \right) }
    = {2\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } }
    = {2\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}} \right)} \right|_0^5} \right] }
    = {2\int\limits_0^{2\pi } {d\varphi } \cdot \frac{{{5^4}}}{4} }
    = {\frac{{625}}{2}\int\limits_0^{2\pi } {d\varphi } }
    = {\frac{{625}}{2} \cdot 2\pi }={ 625\pi .}
    \]

    Page 1
    Problem 1
    Page 2
    Problems 2-5