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Triple Integrals

# Triple Integrals in Spherical Coordinates

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Problem 1
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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.

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
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Problems 2-5