# Calculus

Triple Integrals# Triple Integrals in Cylindrical Coordinates

Problem 1

Problems 2-5

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).

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.}

\]

Figure 1.

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

## Solved Problems

Click on problem description to see solution.

### ✓ Example 1

Evaluate the integral

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

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

### ✓ Example 4

Calculate the integral using cylindrical coordinates:

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

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

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.*

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

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}.}

\]

Figure 2.

Figure 3.

Problem 1

Problems 2-5