Calculus

Double Integrals

Double Integrals in Polar Coordinates

Page 1
Problems 1-2
Page 2
Problems 3-5

One of the particular cases of change of variables is the transformation from Cartesian to polar coordinate system (Figure \(1\)).
\[x = r\cos \theta ,\;\;y = r\sin \theta .\] The Jacobian determinant for this transformation is
\[
{\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}} }
= {\left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial r}}}&{\frac{{\partial x}}{{\partial \theta }}}\\
{\frac{{\partial y}}{{\partial r}}}&{\frac{{\partial y}}{{\partial \theta }}}
\end{array}} \right| }
= {\left| {\begin{array}{*{20}{c}}
{\frac{{\partial \left( {r\cos \theta } \right)}}{{\partial r}}}&{\frac{{\partial \left( {r\cos \theta } \right)}}{{\partial \theta }}}\\
{\frac{{\partial \left( {r\sin \theta } \right)}}{{\partial r}}}&{\frac{{\partial \left( {r\sin \theta } \right)}}{{\partial \theta }}}
\end{array}} \right| }
= {\left| {\begin{array}{*{20}{c}}
{\cos \theta }&{ – r\sin \theta }\\
{\sin\theta }&{r\cos \theta }
\end{array}} \right| }
= {\cos \theta \cdot r\cos \theta }-{ \left( { – r\sin \theta } \right) \cdot \sin \theta }
= {r\,{\cos ^2}\theta + r\,{\sin ^2}\theta }
= {r\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) }={ r.}
\]

Polar coordinate system

Figure 1.

A region of integration in polar coordinates

Figure 2.

Polar rectangle

Figure 3.

As a result, the differential for polar coordinates is
\[
{dxdy }={ \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}}} \right|drd\theta }
= {rdrd\theta .}
\] Let the region \(R\) in polar coordinates be defined as follows (Figure \(2\)):
\[
{0 \le g\left( \theta \right) \le r \le h\left( \theta \right),\;\;}\kern-0.3pt
{\alpha \le \theta \le \beta ,\;\;}\kern-0.3pt
{\text{where}\;\;\beta – \alpha \le 2\pi .}
\] Then the double integral in polar coordinates is given by the formula
\[
{\iint\limits_R {f\left( {x,y} \right)dxdy} }
= {\int\limits_\alpha ^\beta {\int\limits_{g\left( \theta \right)}^{h\left( \theta \right)} {f\left( {r\cos \theta ,r\sin \theta } \right)rdrd\theta } } }
\] The region of integration (Figure \(3\)) is called the polar rectangle if it satisfies the following conditions:
\[
{0 \le a \le r \le b,\;\;}\kern-0.3pt
{\alpha \le \theta \le \beta ,\;\;}\kern-0.3pt
{\text{where}\;\;\beta – \alpha \le 2\pi .}
\] In this case the formula for change of variables can be written as

\[
{\iint\limits_R {f\left( {x,y} \right)dxdy} }
= {\int\limits_\alpha ^\beta {\int\limits_{a}^{b} {f\left( {r\cos \theta ,r\sin \theta } \right)rdrd\theta } } .}
\] Be careful not to forget the factor \(r\) (the Jacobian) in the right-hand side of the formula!

Solved Problems

Click on problem description to see solution.

 Example 1

Calculate the double integral \(\iint\limits_R {\left( {{x^2} + {y^2}} \right)dydx}\) by transforming to polar coordinates. The region of integration \(R\) is the sector \(0 \le \theta \le {\large\frac{\pi }{2}\normalsize}\) of a circle with radius \(r = \sqrt 3 \).

 Example 2

Evaluate the integral \(\iint\limits_R {xydydx},\) where the region of integration \(R\) lies between the circles \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 5.\)

 Example 3

Evaluate the integral \(\iint\limits_R {\sin \theta drd\theta },\) where the region of integration \(R\) is enclosed by the cardioid \(r = 1 + \cos \theta \) (Figure \(6\text{).}\)

 Example 4

Calculate the double integral \(\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy} \) in the circle \({x^2} + {y^2} = 2x.\)

 Example 5

Calculate the double integral \(\iint\limits_R {\sin \sqrt {{x^2} + {y^2}} dxdy} \) by transforming to polar coordinates. The region \(R\) is the disk \({x^2} + {y^2} \le {\pi ^2}.\)

Example 1.

Calculate the double integral \(\iint\limits_R {\left( {{x^2} + {y^2}} \right)dydx}\) by transforming to polar coordinates. The region of integration \(R\) is the sector \(0 \le \theta \le {\large\frac{\pi }{2}\normalsize}\) of a circle with radius \(r = \sqrt 3 \).

Solution.

The region \(R\) is the polar rectangle and described by the set (Figure \(4\))
\[{R }={ \Big\{ {\left( {r,\theta } \right)|\;0 \le r \le \sqrt 3 ,\; }}\kern0pt{{ 0 \le \theta \le {\frac{\pi }{2}}} \Big\} }\] Using the formula
\[
{\iint\limits_R {f\left( {x,y} \right)dxdy} }
= {\int\limits_\alpha ^\beta {\int\limits_{a}^{b} {f\left( {r\cos \theta ,r\sin \theta } \right)rdrd\theta } },}
\] we obtain

Region of integration as a sector of a circle

Figure 4.

\[
{\iint\limits_R {\left( {{x^2} + {y^2}} \right)dydx} }
= {\int\limits_0^{\frac{\pi }{2}} {\int\limits_0^{\sqrt 3 } {{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)rdrd\theta } } }
= {\int\limits_0^{\frac{\pi }{2}} {d\theta } \int\limits_0^{\sqrt 3 } {{r^3}dr} }
= {\left. \theta \right|_0^{\frac{\pi }{2}} \cdot \left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^{\sqrt 3 } }
= {\frac{\pi }{2} \cdot \frac{9}{4} }={ \frac{{9\pi }}{8}.}
\]

Example 2.

Evaluate the integral \(\iint\limits_R {xydydx},\) where the region of integration \(R\) lies between the circles \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 5.\)

The region of integration is a polar rectangle (ring)

Figure 5.

Solution.

In polar coordinates, the region of integration \(R\) is the polar rectangle (Figure \(5\)):
\[{R }={ \left\{ {\left( {r,\theta } \right)|\;1 \le r \le \sqrt 5 ,\;}\right.}\kern0pt {\left.{ 0 \le \theta \le 2\pi } \right\}.}\] So using the formula
\[
{\iint\limits_R {f\left( {x,y} \right)dxdy} }
= {\int\limits_\alpha ^\beta {\int\limits_{a}^{b} {f\left( {r\cos \theta ,r\sin \theta } \right)rdrd\theta } },}
\] we find the integral:

\[
{\iint\limits_R {xydydx} }
= {\int\limits_0^{2\pi } {\int\limits_1^{\sqrt 5 } {r\cos \theta r\sin \theta rdrd\theta } } }
= {\int\limits_0^{2\pi } {\sin \theta \cos \theta d\theta } \int\limits_1^{\sqrt 5 } {{r^3}dr} }
= {\frac{1}{2}\int\limits_0^{2\pi } {\sin 2\theta d\theta } \int\limits_1^{\sqrt 5 } {{r^3}dr} }
= {\frac{1}{2}\left. {\left( { – \frac{{\cos 2\theta }}{2}} \right)} \right|_0^{2\pi } \cdot \left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_1^{\sqrt 5 } }
= {\frac{1}{4}\left( { – \cos 4\pi + \cos 0} \right) \cdot \frac{1}{4}\left( {25 – 1} \right) }
= {\frac{1}{4}\left( { – 1 + 1} \right) \cdot 6 }={ 0.}
\]

Page 1
Problems 1-2
Page 2
Problems 3-5