# Calculus

## Triple Integrals # Triple Integrals in Cylindrical Coordinates

The position of a point $$M\left( {x,y,z} \right)$$ in the $$xyz$$-space in cylindrical coordinates is defined by three numbers: $$\rho, \varphi, z,$$ where $$\rho$$ is the projection of the radius vector of the point $$M$$ onto the $$xy$$-plane, $$\varphi$$ is the angle formed by the projection of the radius vector with the $$x$$-axis (Figure $$1$$), $$z$$ is the projection of the radius vector on the $$z$$-axis (its value is the same in Cartesian and cylindrical coordinates). Figure 1.

The relationship between cylindrical and Cartesian coordinates of a point is given by

${x = \rho \cos \varphi ,\;\;\;}\kern0pt {y = \rho \sin \varphi ,\;\;\;}\kern0pt {z = z.}$

We assume here that

${\rho \ge 0,\;\;\;}\kern-0.3pt {0 \le \varphi \le 2\pi ,\;\;\;}\kern-0.3pt {- \infty \lt z \lt \infty .}$

The Jacobian of transformation from Cartesian to cylindrical coordinates is

${I\left( {\rho ,\varphi ,z} \right) } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial z}}}\\ {\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial z}}}\\ {\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial z}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\cos \varphi }&{ – \rho \sin \varphi }&0\\ {\sin \varphi }&{\rho \cos \varphi }&0\\ 0&0&1 \end{array}} \right| } ={ \rho \ge 0.}$

Then the formula of change of variables for this transformation can be written in the form

${\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} } = {\iiint\limits_{U’} {f\left( {\rho \cos \varphi , }\kern0pt{ \rho \sin \varphi ,z} \right)\rho d\rho d\varphi dz} }$

Transition from cylindrical coordinates makes calculation of triple integrals simpler in those cases when the region of integration is formed by a cylindrical surface.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Evaluate the integral
$\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} }\right.}+{\left.{ {y^4}} \right)dxdydz},$
where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$

### Example 2

Find the integral
$\iiint\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} ,$
where the region $$U$$ is bounded by the surfaces $${x^2} + {y^2} = 3z,$$ $$z = 3$$ (Figure $$4\text{).}$$

### Example 3

Using cylindrical coordinates evaluate the integral
${ \int\limits_{ – 2}^2 {dx} \int\limits_{ – \sqrt {4 – {x^2}} }^{\sqrt {4 – {x^2}} } {dy} \int\limits_0^{4 – {x^2} – {y^2}} {{y^2}dz} .}$

### Example 4

Calculate the integral using cylindrical coordinates:
$\iiint\limits_U {\sqrt {{x^2} + {y^2}} dxdydz} .$
The region $$U$$ is bounded by the paraboloid $$z = 4 – {x^2} – {y^2},$$ by the cylinder $${x^2} + {y^2} = 4$$ and by the planes $$y = 0,$$ $$z = 0$$ (Figure $$8\text{).}$$

### Example 5

Find the integral
$\iiint\limits_U {ydxdydz},$
where the region $$U$$ is bounded by the planes $$z = x + 1,$$ $$z = 0$$ and by the cylindrical surfaces $${x^2} + {y^2} = 1,$$ $${x^2} + {y^2} = 4$$ (see Figure $$10$$).

### Example 1.

Evaluate the integral
$\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} }\right.}+{\left.{ {y^4}} \right)dxdydz},$
where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$

Solution. Figure 2. Figure 3.

It is more convenient to calculate this integral in cylindrical coordinates. Projection of the region of integration onto the $$xy$$-plane is the circle $${x^2} + {y^2} \le 1$$ or $$0 \le \rho \le 1$$ (Figure $$3$$).

Notice that the integrand can be written as

${\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) } = {{\left( {{x^2} + {y^2}} \right)^2} } = {{\left( {{\rho ^2}} \right)^2} = {\rho ^4}.}$

Then the integral becomes

$I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} .$

The second integral contains the factor $$\rho$$ which is the Jacobian of transformation of the Cartesian coordinates into cylindrical coordinates. All the three integrals over each of the variables do not depend on each other. As a result the triple integral is easy to calculate as

${I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} } = {2\pi \int\limits_0^1 {{\rho ^5}d\rho } \int\limits_0^1 {dz} } = {2\pi \cdot 1 \cdot \int\limits_0^1 {{\rho ^5}d\rho } } = {2\pi \left. {\left( {\frac{{{\rho ^6}}}{6}} \right)} \right|_0^1 } = {2\pi \cdot \frac{1}{6} = \frac{\pi }{3}.}$

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Problem 1
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Problems 2-5