One of the particular cases of change of variables is the transformation from Cartesian to polar coordinate system \(\left({\text{Figure }1}\right):\)
\[x = r\cos \theta ,\;\;y = r\sin \theta .\]
Figure 1. 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 \(2\)):
\[0 \le g\left( \theta \right) \le r \le h\left( \theta \right),\;\; \alpha \le \theta \le \beta ,\;\; \text{where}\;\;\beta - \alpha \le 2\pi .\]
Figure 2. Figure 3. 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
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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 \(0 \le \theta \le {\frac{\pi }{2}}\) of a circle with radius \(r = \sqrt 3.\)
Example 2
Evaluate the integral \[\iint\limits_R {xydydx},\] where the region of integration \(R\) lies in the sector \(0 \le \theta \le \frac{\pi}{2}\) between the curves \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 5.\)
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 \(0 \le \theta \le {\frac{\pi }{2}}\) of a circle with radius \(r = \sqrt 3.\)
Solution.
The region \(R\) is the polar rectangle \(\left({\text{Figure }4}\right)\) and described by the set
\[R = \Big\{ {\left( {r,\theta } \right)|\;0 \le r \le \sqrt 3 ,\; 0 \le \theta \le {\frac{\pi }{2}}} \Big\} .\]
Figure 4. 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
\[\iint\limits_R {\left( {{x^2} + {y^2}} \right)dydx} = \int\limits_0^{\frac{\pi }{2}} {\int\limits_0^{\sqrt 3 } {{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)rdrd\theta } } = \int\limits_0^{\frac{\pi }{2}} {d\theta } \int\limits_0^{\sqrt 3 } {{r^3}dr} = \left. \theta \right|_0^{\frac{\pi }{2}} \cdot \left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^{\sqrt 3 } = \frac{\pi }{2} \cdot \frac{9}{4} = \frac{{9\pi }}{8}.\]
Example 2.
Evaluate the integral \[\iint\limits_R {xydydx},\] where the region of integration \(R\) lies in the sector \(0 \le \theta \le \frac{\pi}{2}\) between the curves \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 5.\)
Solution.
In polar coordinates, the region of integration \(R\) is the polar rectangle \(\left({\text{Figure }5}\right):\)
\[R = \left\{ {\left( {r,\theta } \right)|\;1 \le r \le \sqrt 5 ,\; 0 \le \theta \le \frac{\pi}{2} } \right\}.\]
Figure 5. 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:
\[\iint\limits_R {xydydx} = \int\limits_0^{\frac{\pi}{2}} {\int\limits_1^{\sqrt 5 } {r\cos \theta r\sin \theta rdrd\theta } } = \int\limits_0^{\frac{\pi}{2}} {\sin \theta \cos \theta d\theta } \int\limits_1^{\sqrt 5 } {{r^3}dr} = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} {\sin 2\theta d\theta } \int\limits_1^{\sqrt 5 } {{r^3}dr} = \frac{1}{2}\left. {\left( { - \frac{{\cos 2\theta }}{2}} \right)} \right|_0^{\frac{\pi}{2}} \cdot \left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_1^{\sqrt 5 } = \frac{1}{4}\left( { - \cos \pi + \cos 0} \right) \cdot \frac{1}{4}\left( {25 - 1} \right) = \frac{1}{4}\left( { 1 + 1} \right) \cdot 6 = 3.\]
