# Double Integrals in Polar Coordinates

• One of the particular cases of change of variables is the transformation from Cartesian to polar coordinate system $$\left({\text{Figure }1}\right):$$

$x = r\cos \theta ,\;\;y = r\sin \theta .$

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 $$2$$):

${0 \le g\left( \theta \right) \le r \le h\left( \theta \right),\;\;}\kern-0.3pt {\alpha \le \theta \le \beta ,\;\;}\kern-0.3pt {\text{where}\;\;\beta – \alpha \le 2\pi .}$

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:

${0 \le a \le r \le b,\;\;}\kern-0.3pt {\alpha \le \theta \le \beta ,\;\;}\kern-0.3pt {\text{where}\;\;\beta – \alpha \le 2\pi .}$

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 a problem to see the 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 in the sector $$0 \le \theta \le \large{\frac{\pi}{2}}\normalsize$$ between the curves $${x^2} + {y^2} = 1$$ 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 upper half of cardioid $$r = 1 + \cos \theta$$ 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 $${x^2} + {y^2} = 2x.$$

### 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 $$\left({\text{Figure }4}\right)$$ and described by the set

${R }={ \Big\{ {\left( {r,\theta } \right)|\;0 \le r \le \sqrt 3 ,\; }}\kern0pt{{ 0 \le \theta \le {\frac{\pi }{2}}} \Big\} }.$

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

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

### Example 2.

Evaluate the integral $$\iint\limits_R {xydydx},$$ where the region of integration $$R$$ lies in the sector $$0 \le \theta \le \large{\frac{\pi}{2}}\normalsize$$ 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 $$\left({\text{Figure }5}\right):$$

${R }={ \left\{ {\left( {r,\theta } \right)|\;1 \le r \le \sqrt 5 ,\;}\right.}\kern0pt {\left.{ 0 \le \theta \le \frac{\pi}{2} } \right\}.}$

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:

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

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Problems 1-2
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Problems 3-5