Double Integrals in Polar Coordinates

Double Integrals in Polar Coordinates

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

\[
\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}}
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial r}}}&{\frac{{\partial x}}{{\partial \theta }}}\\
{\frac{{\partial y}}{{\partial r}}}&{\frac{{\partial y}}{{\partial \theta }}}
\end{array}} \right|
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial \left( {r\cos \theta } \right)}}{{\partial r}}}&{\frac{{\partial \left( {r\cos \theta } \right)}}{{\partial \theta }}}\\
{\frac{{\partial \left( {r\sin \theta } \right)}}{{\partial r}}}&{\frac{{\partial \left( {r\sin \theta } \right)}}{{\partial \theta }}}
\end{array}} \right|
= \left| {\begin{array}{*{20}{c}}
{\cos \theta }&{ – r\sin \theta }\\
{\sin\theta }&{r\cos \theta }
\end{array}} \right|
= \cos \theta \cdot r\cos \theta – \left( { – r\sin \theta } \right) \cdot \sin \theta
= r\,{\cos ^2}\theta + r\,{\sin ^2}\theta
= r\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = r.\]

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 \(2\)):

\[0 \le g\left( \theta \right) \le r \le h\left( \theta \right),\;\; \alpha \le \theta \le \beta ,\;\; \text{where}\;\;\beta – \alpha \le 2\pi .\]

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:

\[0 \le a \le r \le b,\;\; \alpha \le \theta \le \beta ,\;\; \text{where}\;\;\beta – \alpha \le 2\pi .\]

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

Solution.

The region \(R\) is the polar rectangle \(\left({\text{Figure }4}\right)\) and described by the set

\[R = \Big\{ {\left( {r,\theta } \right)|\;0 \le r \le \sqrt 3 ,\; 0 \le \theta \le {\frac{\pi }{2}}} \Big\} .\]

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

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

Solution.

In polar coordinates, the region of integration \(R\) is the polar rectangle \(\left({\text{Figure }5}\right):\)

\[R = \left\{ {\left( {r,\theta } \right)|\;1 \le r \le \sqrt 5 ,\; 0 \le \theta \le \frac{\pi}{2} } \right\}.\]

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:

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