# Double Integrals in Polar Coordinates

## Solved Problems

Click or tap a problem to see the solution.

### 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 θ and the x-axis (Figure 6).

### Example 4

Calculate the double integral $\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy}$ in the circle x² + y² = 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 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.

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 $${\text{Figure }7.}$$

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 $$8.$$

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,

$I = 2\pi \int\limits_0^\pi {r\sin rdr} = 2\pi \Big[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi - \int\limits_0^\pi {\left( { - \cos r} \right)dr} } \Big] = 2\pi \Big[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi + \int\limits_0^\pi {\cos rdr} } \Big] = 2\pi \left[ {\left. {\left( { - r\cos r} \right)} \right|_0^\pi + \left. {\left( {\sin r} \right)} \right|_0^\pi } \right] = 2\pi \left. {\left( {\sin r - r\cos r} \right)} \right|_0^\pi = 2\pi \left[ {\left( {\sin \pi - \pi \cos \pi } \right) - \left( {\sin 0 - 0 \cdot \cos 0} \right)} \right] = 2\pi \cdot \pi = 2{\pi ^2}.$