Double Integrals in Polar Coordinates
Solved Problems
Example 3.
Evaluate the integral \[\iint\limits_R {\sin \theta drd\theta },\] where the region of integration \(R\) is enclosed by the upper half of cardioid \(r = 1 + \cos \theta \) and the \(x-\)axis (Figure \(6\)).
Solution.
Figure 6. In polar coordinates, the integral is given by
\[\iint\limits_R {\sin \theta drd\theta }
= \int\limits_0^{\pi } {\int\limits_0^{1 + \cos \theta } {\sin \theta drd\theta } }
= \int\limits_0^{\pi } {\left[ {\int\limits_0^{1 + \cos \theta } {dr} } \right]\sin \theta d\theta }
= \int\limits_0^{\pi } {\left[ {\left. r \right|_0^{1 + \cos \theta }} \right]\sin \theta d\theta }
= \int\limits_0^{\pi } {\left( {1 + \cos\theta } \right)\sin \theta d\theta }
= \int\limits_0^{\pi } {\left( {\sin \theta + \cos\theta \sin \theta } \right)d\theta }
= \int\limits_0^{\pi } {\sin \theta d\theta } + \int\limits_0^{\pi } {\frac{{\sin 2\theta }}{2}d\theta }
= \left. {\left( { - \cos \theta } \right)} \right|_0^{\pi } + \frac{1}{2}\left. {\left( { - \frac{{\cos 2\theta }}{2}} \right)} \right|_0^{\pi }
= - \cos \pi + \cos 0 - \frac{1}{4}\cos 4\pi + \frac{1}{4}\cos 0
= 1 + 1 - \cancel{\frac{1}{4}} + \cancel{\frac{1}{4}} = 2.\]
Example 4.
Calculate the double integral \[\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy} \] in the circle \({x^2} + {y^2} = 2x.\)
Solution.
The region of integration \(R\) is shown in \({\text{Figure }7.}\)
Figure 7. We can transform the equation of the circle as
\[{x^2} + {y^2} = 2x,\;\; \Rightarrow {x^2} - 2x + 1 + {y^2} = 1,\;\; \Rightarrow {\left( {x - 1} \right)^2} + {y^2} = 1.\]
Substituting the expressions \(x = r\cos \theta ,\) \(y = r\sin \theta ,\) we obtain the equation of the circle in polar coordinates.
\[{x^2} + {y^2} = 2x,\;\; \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin^2}\theta = 2r\cos \theta ,\;\; \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {\sin^2}\theta } \right) = 2r\cos \theta ,\;\; \Rightarrow r = 2\cos \theta .\]
After transition to polar coordinates we can calculate the double integral:
\[
\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy}
= \iint\limits_S {\left( {{r^2}{{\cos }^2}\theta + {r^2}{\sin^2}\theta } \right)rdrd\theta }
= \iint\limits_S {{r^3}drd\theta }
= \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {\int\limits_0^{2\cos \theta } {{r^3}dr} } \right]d\theta }
= 4\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^{2\cos \theta }} \right]d\theta }
= 4\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^4}\theta d\theta }
= 4\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\left( {\frac{{1 + \cos 2\theta }}{2}} \right)}^2}d\theta }
= \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + 2\cos 2\theta + {{\cos }^2}2\theta } \right)d\theta }
= \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {\frac{3}{2} + 2\cos 2\theta + \frac{1}{2}\cos 4\theta } \right)d\theta }
= \left. {\left( {\frac{3}{2}\theta + \sin 2\theta + \frac{1}{8}\sin 4\theta } \right)} \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}
= \left( {\frac{3}{2} \cdot \frac{\pi }{2} + \sin \pi + \frac{1}{8}\sin 2\pi } \right) - \left( { - \frac{3}{2} \cdot \frac{\pi }{2} - \sin \pi - \frac{1}{8}\sin 2\pi } \right)
= \frac{{3\pi }}{2}.\]
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 \({x^2} + {y^2} \le {\pi ^2}.\)
Solution.
The region \(R\) is presented in Figure \(8.\)
Figure 8. Figure 9. The image \(S\) of the initial region \(R\) is defined by the set
\[\left\{ {S = \left( {r,\theta } \right)|\;0 \le r \le \pi ,\; 0 \le \theta \le 2\pi } \right\} .\]
and is shown in Figure \(9.\) The double integral in polar coordinates becomes
\[I = \iint\limits_R {\sin \sqrt {{x^2} + {y^2}} dxdy} = \iint\limits_S {r\sin rdrd\theta } = \int\limits_0^{2\pi } {d\theta } \int\limits_0^\pi {r\sin rdr} = 2\pi \int\limits_0^\pi {r\sin rdr} .\]
We compute this integral using integration by parts:
\[\int\limits_a^b {udv} = \left. {\left( {uv} \right)} \right|_a^b - \int\limits_a^b {vdu} .\]
Let \(u = r,\) \(dv = \sin rdr.\) Then \(du = dr,\) \(v = \int {\sin rdr} \) \(= - \cos r.\) Hence,
\[
I = 2\pi \int\limits_0^\pi {r\sin rdr}
= 2\pi \Big[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi - \int\limits_0^\pi {\left( { - \cos r} \right)dr} } \Big] = 2\pi \Big[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi + \int\limits_0^\pi {\cos rdr} } \Big]
= 2\pi \left[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi + \left. {\left( {\sin r} \right)} \right|_0^\pi } \right]
= 2\pi \left. {\left( {\sin r - r\cos r} \right)} \right|_0^\pi
= 2\pi \left[ {\left( {\sin \pi - \pi \cos \pi } \right) - \left( {\sin 0 - 0 \cdot \cos 0} \right)} \right]
= 2\pi \cdot \pi = 2{\pi ^2}.\]
