15.3: Double Integrals in Polar Coordinates
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Recognize the format of a double integral over a polar rectangular region.
- Evaluate a double integral in polar coordinates by using an iterated integral.
- Recognize the format of a double integral over a general polar region.
- Use double integrals in polar coordinates to calculate areas and volumes.
Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.
Polar Rectangular Regions of Integration
When we defined the double integral for a continuous function in rectangular coordinates—say, g over a region R in the xy-plane—we divided R into subrectangles with sides parallel to the coordinate axes. These sides have either constant x-values and/or constant y-values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant r-values and/or constant θ-values. This means we can describe a polar rectangle as in Figure 15.3.1a, with R={(r,θ)|a≤r≤b,α≤θ≤β}.

In this section, we are looking to integrate over polar rectangles. Consider a function f(r,θ) over a polar rectangle R. We divide the interval [a,b] into m subintervals [ri−1,ri] of length Δr=(b−a)/m and divide the interval [α,β] into n subintervals [θi−1,θi] of width Δθ=(β−α)/n. This means that the circles r=ri and rays θ=θi for 1≤i≤m and 1≤j≤n divide the polar rectangle R into smaller polar subrectangles Rij (Figure 15.3.1b).
As before, we need to find the area ΔA of the polar subrectangle Rij and the “polar” volume of the thin box above Rij. Recall that, in a circle of radius r the length s of an arc subtended by a central angle of θ radians is s=rθ. Notice that the polar rectangle Rij looks a lot like a trapezoid with parallel sides ri−1Δθ and riΔθ and with a width Δr. Hence the area of the polar subrectangle Rij is
ΔA=12Δr(ri−1Δθ+riΔθ).
Simplifying and letting
r∗ij=12(ri−1+ri)
we have ΔA=r∗ijΔrΔθ.
Therefore, the polar volume of the thin box above Rij (Figure 15.3.2) is
f(r∗ij,θ∗ij)ΔA=f(r∗ij,θ∗ij)r∗ijΔrΔθ.

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as
m∑i=1n∑j=1f(r∗ij,θ∗ij)r∗ijΔrΔθ.
As we have seen before, we obtain a better approximation to the polar volume of the solid above the region R when we let m and n become larger. Hence, we define the polar volume as the limit of the double Riemann sum,
V=limm,n→∞m∑i=1n∑j=1f(r∗ij,θ∗ij)r∗ijΔrΔθ.
This becomes the expression for the double integral.
The double integral of the function f(r,θ) over the polar rectangular region R in the rθ-plane is defined as
∬
Again, just as in section on Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,
\iint_R f(r, \theta)\,dA = \iint_R f(r, \theta) \,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=a}^{r=b} f(r,\theta) \,r \, dr \, d\theta. \nonumber
Notice that the expression for dA is replaced by r \, dr \, d\theta when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function f is given in terms of x and y using x = r \, \cos \, \theta, \, y = r \, \sin \, \theta, and dA = r \, dr \, d\theta changes it to
\iint_R f(x,y) \,dA = \iint_R f(r \, \cos \, \theta, \, r \, \sin \, \theta ) \,r \, dr \, d\theta. \nonumber
Note that all the properties listed in section on Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.
Sketch the polar rectangular region
R = \{(r, \theta)\,|\,1 \leq r \leq 3, 0 \leq \theta \leq \pi \}. \nonumber
Solution
As we can see from Figure \PageIndex{3}, r = 1 and r = 3 are circles of radius 1 and 3 and 0 \leq \theta \leq \pi covers the entire top half of the plane. Hence the region R looks like a semicircular band.

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.
Evaluate the integral \displaystyle \iint_R 3x \, dA over the region R = \{(r, \theta)\,|\,1 \leq r \leq 2, \, 0 \leq \theta \leq \pi \}.
Solution
First we sketch a figure similar to Figure \PageIndex{3}, but with outer radius r=2. From the figure we can see that we have
\begin{align*} \iint_R 3x \, dA &= \int_{\theta=0}^{\theta=\pi} \int_{r=1}^{r=2} 3r \, \cos \, \theta \,r \, dr \, d\theta \quad\text{Use an integral with correct limits of integration.} \\ &= \int_{\theta=0}^{\theta=\pi} \cos \, \theta \left[\left. r^3\right|_{r=1}^{r=2}\right] d\theta \quad\text{Integrate first with respect to $r$.} \\ &=\int_{\theta=0}^{\theta=\pi} 7 \, \cos \, \theta \, d\theta \\ &= 7 \, \sin \, \theta \bigg|_{\theta=0}^{\theta=\pi} = 0. \end{align*}
Sketch the region D = \{ (r,\theta) \vert 1\leq r \leq 2, \, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \}, and evaluate \displaystyle \iint_R x \, dA.
- Hint
-
Follow the steps in Example \PageIndex{1A}.
- Answer
-
\frac{14}{3}
Evaluate the integral
\iint_R (1 - x^2 - y^2) \,dA \nonumber
where R is the unit circle on the xy-plane.
Solution
The region R is a unit circle, so we can describe it as R = \{(r, \theta )\,|\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi \}.
Using the conversion x = r \, \cos \, \theta, \, y = r \, \sin \, \theta, and dA = r \, dr \, d\theta, we have
\begin{align*} \iint_R (1 - x^2 - y^2) \,dA &= \int_0^{2\pi} \int_0^1 (1 - r^2) \,r \, dr \, d\theta \\[4pt] &= \int_0^{2\pi} \int_0^1 (r - r^3) \,dr \, d\theta \\ &= \int_0^{2\pi} \left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 \,d\theta \\&= \int_0^{2\pi} \frac{1}{4}\,d\theta = \frac{\pi}{2}. \end{align*}
Evaluate the integral \displaystyle \iint_R (x + y) \,dA \nonumber where R = \big\{(x,y)\,|\,1 \leq x^2 + y^2 \leq 4, \, x \leq 0 \big\}.
Solution
We can see that R is an annular region that can be converted to polar coordinates and described as R = \left\{(r, \theta)\,|\,1 \leq r \leq 2, \, \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \right\} (see the following graph).

Hence, using the conversion x = r \, \cos \, \theta, \, y = r \, \sin \, \theta, and dA = r \, dr \, d\theta, we have
\begin{align*} \iint_R (x + y)\,dA &= \int_{\theta=\pi/2}^{\theta=3\pi/2} \int_{r=1}^{r=2} (r \, \cos \, \theta + r \, \sin \, \theta) r \, dr \, d\theta \\ &= \left(\int_{r=1}^{r=2} r^2 \, dr\right)\left(\int_{\pi/2}^{3\pi/2} (\cos \, \theta + \sin \, \theta)\,d\theta\right) \\ &= \left. \left[\frac{r^3}{3}\right]_1^2 [\sin \, \theta - \cos \, \theta] \right|_{\pi/2}^{3\pi/2} \\ &= - \frac{14}{3}. \end{align*}
Evaluate the integral \displaystyle \iint_R (4 - x^2 - y^2)\,dA \nonumber where R is the circle of radius 2 on the xy-plane.
- Hint
-
Follow the steps in the previous example.
- Answer
-
8\pi
General Polar Regions of Integration
To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in section on Double Integrals over General Regions. It is more common to write polar equations as r = f(\theta) than \theta = f(r), so we describe a general polar region as R = \{(r, \theta)\,|\,\alpha \leq \theta \leq \beta, \, h_1 (\theta) \leq r \leq h_2(\theta)\} (Figure \PageIndex{5}).

If f(r, \theta) is continuous on a general polar region D as described above, then
\iint_D f(r, \theta ) \,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=h_1(\theta)}^{r=h_2(\theta)} f(r,\theta) \, r \, dr \, d\theta. \nonumber
Evaluate the integral
\iint_D r^2 \sin \theta \, r \, dr \, d\theta \nonumber
where D is the region bounded by the polar axis and the upper half of the cardioid r = 1 + \cos \, \theta.
Solution
We can describe the region D as \{(r, \theta)\,|\,0 \leq \theta \leq \pi, \, 0 \leq r \leq 1 + \cos \, \theta\} as shown in Figure \PageIndex{6}.

Hence, we have
\begin{align*} \iint_D r^2 \sin \, \theta \, r \, dr \, d\theta &= \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=1+\cos \theta} (r^2 \sin \, \theta) \,r \, dr \, d\theta \\ &= \frac{1}{4}\left.\int_{\theta=0}^{\theta=\pi}[r^4] \right|_{r=0}^{r=1+\cos \, \theta} \sin \, \theta \, d\theta \\ &= \frac{1}{4} \int_{\theta=0}^{\theta=\pi} (1 + \cos \, \theta )^4 \sin \, \theta \, d\theta \\ &= - \frac{1}{4} \left[ \frac{(1 + \cos \, \theta)^5}{5}\right]_0^{\pi} = \frac{8}{5}.\end{align*}
Evaluate the integral
\iint_D r^2 \sin^2 2\theta \,r \, dr \, d\theta \nonumber
where D = \left\{ (r,\theta)\,|\,0 \leq \theta \leq \pi, \, 0 \leq r \leq 2 \sqrt{\cos \, 2\theta} \right\}.
- Hint
-
Graph the region and follow the steps in the previous example.
- Answer
-
\frac{\pi}{8}
Polar Areas and Volumes
As in rectangular coordinates, if a solid S is bounded by the surface z = f(r, \theta), as well as by the surfaces r = a, \, r = b, \, \theta = \alpha, and \theta = \beta, we can find the volume V of S by double integration, as
V = \iint_R f(r, \theta) \,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=a}^{r=b} f(r,\theta)\, r \, dr \, d\theta. \nonumber
If the base of the solid can be described as D = \{(r, \theta)|\alpha \leq \theta \leq \beta, \, h_1 (\theta) \leq r \leq h_2(\theta)\}, then the double integral for the volume becomes
V = \iint_D f(r, \theta) \,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=h_1(\theta)}^{r=h_2(\theta)} f(r,\theta) \,r \, dr \, d\theta. \nonumber
We illustrate this idea with some examples.
Find the volume of the solid that lies under the paraboloid z = 1 - x^2 - y^2 and above the unit circle on the xy-plane (Figure \PageIndex{7}).

Solution
By the method of double integration, we can see that the volume is the iterated integral of the form
\displaystyle \iint_R (1 - x^2 - y^2)\,dA \nonumber
where R = \big\{(r, \theta)\,|\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi\big\}.
This integration was shown before in Example \PageIndex{2A}, so the volume is \frac{\pi}{2} cubic units.
Find the volume of the solid that lies under the paraboloid z = 4 - x^2 - y^2 and above the disk (x - 1)^2 + y^2 = 1 on the xy-plane. See the paraboloid in Figure \PageIndex{8} intersecting the cylinder (x - 1)^2 + y^2 = 1 above the xy-plane.

Solution
First change the disk (x - 1)^2 + y^2 = 1 to polar coordinates. Expanding the square term, we have x^2 - 2x + 1 + y^2 = 1. Then simplify to get x^2 + y^2 = 2x, which in polar coordinates becomes r^2 = 2r \, \cos \, \theta and then either r = 0 or r = 2 \, \cos \, \theta. Similarly, the equation of the paraboloid changes to z = 4 - r^2. Therefore we can describe the disk (x - 1)^2 + y^2 = 1 on the xy -plane as the region
D = \{(r,\theta)\,|\,0 \leq \theta \leq \pi, \, 0 \leq r \leq 2 \, \cos \theta\}. \nonumber
Hence the volume of the solid bounded above by the paraboloid z = 4 - x^2 - y^2 and below by r = 2 \, \cos \theta is
\begin{align*} V &= \iint_D f(r, \theta) \,r \, dr \, d\theta \\&= \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=2 \, \cos \, \theta} (4 - r^2) \,r \, dr \, d\theta\\ &= \int_{\theta=0}^{\theta=\pi}\left.\left[4\frac{r^2}{2} - \frac{r^4}{4}\right|_0^{2 \, \cos \, \theta}\right]d\theta \\ &= \int_0^{\pi} [8 \, \cos^2\theta - 4 \, \cos^4\theta]\,d\theta \\&= \left[\frac{5}{2}\theta + \frac{5}{2} \sin \, \theta \, \cos \, \theta - \sin \, \theta \cos^3\theta \right]_0^{\pi} = \frac{5}{2}\pi\; \text{units}^3. \end{align*}
Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if f has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.
Find the volume of the region that lies under the paraboloid z = x^2 + y^2 and above the triangle enclosed by the lines y = x, \, x = 0, and x + y = 2 in the xy-plane.
Solution
First examine the region over which we need to set up the double integral and the accompanying paraboloid.

The region D is \{(x,y)\,|\,0 \leq x \leq 1, \, x \leq y \leq 2 - x\}. Converting the lines y = x, \, x = 0, and x + y = 2 in the xy-plane to functions of r and \theta we have \theta = \pi/4, \, \theta = \pi/2, and r = 2 / (\cos \, \theta + \sin \, \theta), respectively. Graphing the region on the xy- plane, we see that it looks like D = \{(r, \theta)\,|\,\pi/4 \leq \theta \leq \pi/2, \, 0 \leq r \leq 2/(\cos \, \theta + \sin \, \theta)\}.
Now converting the equation of the surface gives z = x^2 + y^2 = r^2. Therefore, the volume of the solid is given by the double integral
\begin{align*} V &= \iint_D f(r, \theta)\,r \, dr \, d\theta \\&= \int_{\theta=\pi/4}^{\theta=\pi/2} \int_{r=0}^{r=2/ (\cos \, \theta + \sin \, \theta)} r^2 r \, dr d\theta \\ &= \int_{\pi/4}^{\pi/2}\left[\frac{r^4}{4}\right]_0^{2/(\cos \, \theta + \sin \, \theta)} d\theta \\ &=\frac{1}{4}\int_{\pi/4}^{\pi/2} \left(\frac{2}{\cos \, \theta + \sin \, \theta}\right)^4 d\theta \\ &= \frac{16}{4} \int_{\pi/4}^{\pi/2} \left(\frac{1}{\cos \, \theta + \sin \, \theta} \right)^4 d\theta \\&= 4\int_{\pi/4}^{\pi/2} \left(\frac{1}{\cos \, \theta + \sin \, \theta}\right)^4 d\theta. \end{align*}
As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as
V = \int_0^1 \int_x^{2-x} (x^2 + y^2) \,dy \, dx. \nonumber
Evaluating gives
\begin{align*} V &= \int_0^1 \int_x^{2-x} (x^2 + y^2) \,dy \, dx \\&= \int_0^1 \left.\left[x^2y + \frac{y^3}{3}\right]\right|_x^{2-x} dx\\ &= \int_0^1 \frac{8}{3} - 4x + 4x^2 - \frac{8x^3}{3} \,dx \\ &= \left.\left[\frac{8x}{3} - 2x^2 + \frac{4x^3}{3} - \frac{2x^4}{3}\right]\right|_0^1 \\&= \frac{4}{3} \; \text{units}^3. \end{align*}
To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.
Use polar coordinates to find the volume inside the cone z = 2 - \sqrt{x^2 + y^2} and above the xy-plane.
Solution
The region D for the integration is the base of the cone, which appears to be a circle on the xy-plane (Figure \PageIndex{10}).

We find the equation of the circle by setting z = 0:
\begin{align*} 0 &= 2 - \sqrt{x^2 + y^2} \\ 2 &= \sqrt{x^2 + y^2} \\ x^2 + y^2 &= 4. \end{align*}
This means the radius of the circle is 2 so for the integration we have 0 \leq \theta \leq 2\pi and 0 \leq r \leq 2. Substituting x = r \, \cos \theta and y = r \, \sin \, \theta in the equation z = 2 - \sqrt{x^2 + y^2} we have z = 2 - r. Therefore, the volume of the cone is
\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=2} (2 - r)\,r \, dr \, d\theta = 2 \pi \frac{4}{3} = \frac{8\pi}{3}\; \text{cubic units.} \nonumber
AnalysisNote that if we were to find the volume of an arbitrary cone with radius \alpha units and height h units, then the equation of the cone would be z = h - \frac{h}{a}\sqrt{x^2 + y^2}.
We can still use Figure \PageIndex{10} and set up the integral as
\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=a} \left(h - \frac{h}{a}r\right) r \, dr \, d\theta. \nonumber
Evaluating the integral, we get \frac{1}{3} \pi a^2 h.
Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids z = x^2 + y^2 and z = 16 - x^2 - y^2.
- Hint
-
Sketching the graphs can help.
- Answer
-
V = \int_0^{2\pi} \int_0^{2\sqrt{2}} (16 - 2r^2) \,r \, dr \, d\theta = 64 \pi \; \text{cubic units.} \nonumber
Find the area enclosed by the circle r = 3 \, \cos \, \theta and the cardioid r = 1 + \cos \, \theta.
Solution
First and foremost, sketch the graphs of the region (Figure \PageIndex{12}).

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives
3 \, \cos \, \theta = 1 + \cos \, \theta. \nonumber
One of the points of intersection is \theta = \pi/3. The area above the polar axis consists of two parts, with one part defined by the cardioid from \theta = 0 to \theta = \pi/3 and the other part defined by the circle from \theta = \pi/3 to \theta = \pi/2. By symmetry, the total area is twice the area above the polar axis. Thus, we have
A = 2 \left[\int_{\theta=0}^{\theta=\pi/3} \int_{r=0}^{r=1+\cos \, \theta} 1 \,r \, dr \, d\theta + \int_{\theta=\pi/3}^{\theta=\pi/2} \int_{r=0}^{r=3 \, \cos \, \theta} 1\,r \, dr \, d\theta \right]. \nonumber
Evaluating each piece separately, we find that the area is
A = 2 \left(\frac{1}{4}\pi + \frac{9}{16} \sqrt{3} + \frac{3}{8} \pi - \frac{9}{16} \sqrt{3} \right) = 2 \left(\frac{5}{8}\pi\right) = \frac{5}{4}\pi \, \text{square units.} \nonumber
Find the area enclosed inside the cardioid r = 3 - 3 \, \sin \theta and outside the cardioid r = 1 + \sin \theta.
- Hint
-
Sketch the graph, and solve for the points of intersection.
- Answer
-
A = 2 \int_{-\pi/2}^{\pi/6} \int_{1+\sin \, \theta}^{3-3\sin \, \theta} \,r \, dr \, d\theta = \left(8 \pi + 9 \sqrt{3}\right) \; \text{units}^2 \nonumber
Evaluate the integral
\iint_{R^2} e^{-10(x^2+y^2)} \,dx \, dy. \nonumber
Solution
This is an improper integral because we are integrating over an unbounded region R^2. In polar coordinates, the entire plane R^2 can be seen as 0 \leq \theta \leq 2\pi, \, 0 \leq r \leq \infty.
Using the changes of variables from rectangular coordinates to polar coordinates, we have
\begin{align*} \iint_{R^2} e^{-10(x^2+y^2)}\,dx \, dy &= \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=\infty} e^{-10r^2}\,r \, dr \, d\theta = \int_{\theta=0}^{\theta=2\pi} \left(\lim_{a\rightarrow\infty} \int_{r=0}^{r=a} e^{-10r^2}r \, dr \right) d\theta \\ &=\left(\int_{\theta=0}^{\theta=2\pi}\right) d\theta \left(\lim_{a\rightarrow\infty} \int_{r=0}^{r=a} e^{-10r^2}r \, dr \right) \\ &=2\pi \left(\lim_{a\rightarrow\infty} \int_{r=0}^{r=a} e^{-10r^2}r \, dr \right) \\ &=2\pi \lim_{a\rightarrow\infty}\left(-\frac{1}{20}\right)\left(\left. e^{-10r^2}\right|_0^a\right) \\ &=2\pi \left(-\frac{1}{20}\right)\lim_{a\rightarrow\infty}\left(e^{-10a^2} - 1\right) \\ &= \frac{\pi}{10}. \end{align*}
Evaluate the integral
\iint_{R^2} e^{-4(x^2+y^2)}dx \, dy. \nonumber
- Hint
-
Convert to the polar coordinate system.
- Answer
-
\frac{\pi}{4}
Key Concepts
- To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
- The area dA in polar coordinates becomes r \, dr \, d\theta.
- Use x = r \, \cos \, \theta, \, y = r \, \sin \, \theta, and dA = r \, dr \, d\theta to convert an integral in rectangular coordinates to an integral in polar coordinates.
- Use r^2 = x^2 + y^2 and \theta = tan^{-1} \left(\frac{y}{x}\right) to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
- To find the volume in polar coordinates bounded above by a surface z = f(r, \theta) over a region on the xy-plane, use a double integral in polar coordinates.
Key Equations
- Double integral over a polar rectangular region R
\iint_R f(r, \theta) dA = \lim_{m,n\rightarrow\infty}\sum_{i=1}^m \sum_{j=1}^n f(r_{ij}^*, \theta_{ij}^*) \Delta A = \lim_{m,n\rightarrow\infty}\sum_{i=1}^m \sum_{j=1}^nf(r_{ij}^*,\theta_{ij}^*)r_{ij}^*\Delta r \Delta \theta \nonumber
- Double integral over a general polar region
\iint_D f(r, \theta)\,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=h_1(\theta)}^{r_2(\theta)} f (r,\theta) \,r \, dr \, d\theta \nonumber
Glossary
- polar rectangle
- the region enclosed between the circles r = a and r = b and the angles \theta = \alpha and \theta = \beta; it is described as R = \{(r, \theta)\,|\,a \leq r \leq b, \, \alpha \leq \theta \leq \beta\}
As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like
\text{Area of} \, A = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} 1 \,r \, dr \, d\theta. \nonumber
Example \PageIndex{6A}: Finding an Area Using a Double Integral in Polar Coordinates
Evaluate the area bounded by the curve r = \cos \, 4\theta.
Solution
Sketching the graph of the function r = \cos \, 4\theta reveals that it is a polar rose with eight petals (see the following figure).
Using symmetry, we can see that we need to find the area of one petal and then multiply it by 8. Notice that the values of \theta for which the graph passes through the origin are the zeros of the function \cos \, 4\theta, and these are odd multiples of \pi/8. Thus, one of the petals corresponds to the values of \theta in the interval [-\pi/8, \pi/8]. Therefore, the area bounded by the curve r = \cos \, 4\theta is
\begin{align*} A &= 8 \int_{\theta=-\pi/8}^{\theta=\pi/8} \int_{r=0}^{r=\cos \, 4\theta} 1\,r \, dr \, d\theta \\ &= 8 \int_{\theta=-\pi/8}^{\theta=\pi/8}\left.\left[\frac{1}{2}r^2\right|_0^{\cos \, 4\theta}\right] d\theta \\ &= 8 \int_{-\pi/8}^{\pi/8} \frac{1}{2} \cos^24\theta \, d\theta \\&= 8\left. \left[\frac{1}{4} \theta + \frac{1}{16} \sin \, 4\theta \, \cos \, 4\theta \right|_{-\pi/8}^{\pi/8}\right] \\&= 8 \left[\frac{\pi}{16}\right] = \frac{\pi}{2}\; \text{units}^2. \end{align*}