Section 1.4Area and Arc Length in Polar Coordinates

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function \(y=f(x)\) defined from \(x=a\) to \(x=b\) where \(f(x)>0\) on this interval, the area between the curve and the x-axis is given by \(A={∫}_{a}^{b}f(x)\,dx.\) This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by \(L={∫}_{a}^{b}\sqrt{1+{({f}^{\prime }(x))}^{2}}dx.\) In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function \(r=f(θ),\) where \(α\le θ\le β.\) Our first step is to partition the interval \([α,β]\) into n equal-width subintervals. The width of each subinterval is given by the formula \(\Delta θ=(β-α)\text{/}n,\) and the ith partition point \({θ}_{i}\) is given by the formula \({θ}_{i}=α+i\Delta θ.\) Each partition point \(θ={θ}_{i}\) defines a line with slope \(\text{tan}{θ}_{i}\) passing through the pole as shown in the following graph.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

Recall that the area of a circle is \(A=\pi {r}^{2}.\) When measuring angles in radians, 360 degrees is equal to \(2\pi\) radians. Therefore a fraction of a circle can be measured by the central angle \(θ.\) The fraction of the circle is given by \(\frac{θ}{2\pi },\) so the area of the sector is this fraction multiplied by the total area:

\[A=(\frac{θ}{2\pi })\,\pi {r}^{2}=\frac{1}{2}θ{r}^{2}.\]

Since the radius of a typical sector in Figure 1 is given by \({r}_{i}=f({θ}_{i}),\) the area of the ith sector is given by

\[{A}_{i}=\frac{1}{2}(\Delta θ){(f({θ}_{i}))}^{2}.\]

Therefore a Riemann sum that approximates the area is given by

\[{A}_{n}=∑i=1n{A}_{i}\approx ∑i=1n\frac{1}{2}(\Delta θ){(f({θ}_{i}))}^{2}.\]

We take the limit as \(n\to \infty\) to get the exact area:

\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{1}{2}{∫}_{α}^{β}{(f(θ))}^{2}dθ.\]

This gives the following theorem.

Example 1

Find the area of one petal of the rose defined by the equation \(r=3\,\text{sin}(2θ).\)

Find the consecutive θ values where \(r=0\) to locate one petal's bounds, then apply \(A=\frac{1}{2}{∫}_{α}^{β}{[f(θ)]}^{2}dθ\) over that interval.

Show solution

The graph of \(r=3\,\text{sin}(2θ)\) follows.

When \(θ=0\) we have \(r=3\,\text{sin}(2(0))=0.\) The next value for which \(r=0\) is \(θ=\pi \text{/}2.\) This can be seen by solving the equation \(3\,\text{sin}(2θ)=0\) for \(θ.\) Therefore the values \(θ=0\) to \(θ=\pi \text{/}2\) trace out the first petal of the rose. To find the area inside this petal, use Equation 5 with \(f(θ)=3\,\text{sin}(2θ),\) \(α=0,\) and \(β=\pi \text{/}2\text{:}\)

\[\begin{array}{ll}A & =\frac{1}{2}{∫}_{α}^{β}{[f(θ)]}^{2}dθ \\ & =\frac{1}{2}{∫}_{0}^{\pi \text{/}2}{[3\,\text{sin}(2θ)]}^{2}dθ \\ & =\frac{1}{2}{∫}_{0}^{\pi \text{/}2}9\,{\text{sin}}^{2}(2θ)\,dθ.\end{array}\]

To evaluate this integral, use the formula \({\text{sin}}^{2}α=(1-\text{cos}(2α))\text{/}2\) with \(α=2θ\text{:}\)

\[\begin{array}{ll}A & =\frac{1}{2}{∫}_{0}^{\pi \text{/}2}9\,{\text{sin}}^{2}(2θ)\,dθ \\ & =\frac{9}{2}{∫}_{0}^{\pi \text{/}2}\frac{(1-\text{cos}(4θ))}{2}dθ \\ & =\frac{9}{4}({∫}_{0}^{\pi \text{/}2}1-\text{cos}(4θ)\,dθ) \\ & =\frac{9}{4}{(θ-\frac{\text{sin}(4θ)}{4})}_{0}^{\pi \text{/}2} \\ & =\frac{9}{4}(\frac{\pi }{2}-\frac{\text{sin}\,2\pi }{4})-\frac{9}{4}(0-\frac{\text{sin}\,4(0)}{4}) \\ & =\frac{9\pi }{8}.\end{array}\]

Example 1 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Example 2

Find the area outside the cardioid \(r=2+2\,\text{sin}\,θ\) and inside the circle \(r=6\,\text{sin}\,θ.\)

Set the two r-expressions equal to find the θ bounds of intersection, then subtract the cardioid's area from the circle's area over that interval.

Show solution

First draw a graph containing both curves as shown.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for \(θ\text{:}\)

\[\begin{array}{lll}6\,\text{sin}\,θ & = & 2+2\,\text{sin}\,θ \\ 4\,\text{sin}\,θ & = & 2 \\ \text{sin}\,θ & = & \frac{1}{2}.\end{array}\]

This gives the solutions \(θ=\frac{\pi }{6}\) and \(θ=\frac{5\pi }{6},\) which are the limits of integration. The circle \(r=3\,\text{sin}\,θ\) is the red graph, which is the outer function, and the cardioid \(r=2+2\,\text{sin}\,θ\) is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between \(θ=\frac{\pi }{6}\) and \(θ=\frac{5\pi }{6},\) then subtract the area inside the cardioid between \(θ=\frac{\pi }{6}\) and \(θ=\frac{5\pi }{6}\text{:}\)

\[\begin{array}{ll}A & =\text{circle}-\text{cardioid} \\ & =\frac{1}{2}{∫}_{\pi \text{/}6}^{5\pi \text{/}6}{[6\,\text{sin}\,θ]}^{2}dθ-\frac{1}{2}{∫}_{\pi \text{/}6}^{5\pi \text{/}6}{[2+2\,\text{sin}\,θ]}^{2}dθ \\ & =\frac{1}{2}{∫}_{\pi \text{/}6}^{5\pi \text{/}6}36\,{\text{sin}}^{2}θ\,dθ-\frac{1}{2}{∫}_{\pi \text{/}6}^{5\pi \text{/}6}[4+8\,\text{sin}\,θ+4\,{\text{sin}}^{2}θ]\,dθ \\ & =18{∫}_{\pi \text{/}6}^{5\pi \text{/}6}\frac{1-\text{cos}(2θ)}{2}dθ-2{∫}_{\pi \text{/}6}^{5\pi \text{/}6}[1+2\,\text{sin}\,θ+\frac{1-\text{cos}(2θ)}{2}]dθ \\ & =9{[θ-\frac{\text{sin}(2θ)}{2}]}_{\pi \text{/}6}^{5\pi \text{/}6}-2{[\frac{3θ}{2}-2\,\text{cos}\,θ-\frac{\text{sin}(2θ)}{4}]}_{\pi \text{/}6}^{5\pi \text{/}6} \\ & =9(\frac{5\pi }{6}-\frac{\text{sin}\,2(5\pi \text{/}6)}{2})-9(\frac{\pi }{6}-\frac{\text{sin}\,2(\pi \text{/}6)}{2}) \\ & \,\text{-}(3(\frac{5\pi }{6})-4\,\text{cos}\,\frac{5\pi }{6}-\frac{\text{sin}\,2(5\pi \text{/}6)}{2})+(3(\frac{\pi }{6})-4\,\text{cos}\,\frac{\pi }{6}-\frac{\text{sin}\,2(\pi \text{/}6)}{2}) \\ & =4\pi .\end{array}\]

In Example 2 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for \(θ\) yielded two solutions: \(θ=\frac{\pi }{6}\) and \(θ=\frac{5\pi }{6}.\) However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of \(θ.\) For example, for the cardioid we get

\[\begin{array}{lll}2+2\,\text{sin}\,θ & = & 0 \\ \text{sin}\,θ & = & -1,\end{array}\]

so the values for \(θ\) that solve this equation are \(θ=\frac{3\pi }{2}+2n\pi ,\) where n is any integer. For the circle we get

\[6\,\text{sin}\,θ=0.\]

The solutions to this equation are of the form \(θ=n\pi\) for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc Length in Polar Curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve \((x(t),y(t))\) for \(a\le t\le b\) is given by

\[L={∫}_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt.\]

In polar coordinates we define the curve by the equation \(r=f(θ),\) where \(α\le θ\le β.\) In order to adapt the arc length formula for a polar curve, we use the equations

\[x=r\,\text{cos}\,θ=f(θ)\,\text{cos}\,θ\,\text{and}\,y=r\,\text{sin}\,θ=f(θ)\,\text{sin}\,θ,\]

and we replace the parameter t by \(θ.\) Then

\[\begin{array}{l}\frac{dx}{dθ}={f}^{\prime }(θ)\,\text{cos}\,θ-f(θ)\,\text{sin}\,θ \\ \frac{dy}{dθ}={f}^{\prime }(θ)\,\text{sin}\,θ+f(θ)\,\text{cos}\,θ.\end{array}\]

We replace \(dt\) by \(dθ,\) and the lower and upper limits of integration are \(α\) and \(β,\) respectively. Then the arc length formula becomes

\[\begin{array}{ll}L & ={∫}_{a}^{b}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt \\ & ={∫}_{α}^{β}\sqrt{{(\frac{dx}{dθ})}^{2}+{(\frac{dy}{dθ})}^{2}}dθ \\ & ={∫}_{α}^{β}\sqrt{{({f}^{\prime }(θ)\,\text{cos}\,θ-f(θ)\,\text{sin}\,θ)}^{2}+{({f}^{\prime }(θ)\,\text{sin}\,θ+f(θ)\,\text{cos}\,θ)}^{2}}dθ \\ & ={∫}_{α}^{β}\sqrt{{({f}^{\prime }(θ))}^{2}({\text{cos}}^{2}θ+{\text{sin}}^{2}θ)+{(f(θ))}^{2}({\text{cos}}^{2}θ+{\text{sin}}^{2}θ)}dθ \\ & ={∫}_{α}^{β}\sqrt{{({f}^{\prime }(θ))}^{2}+{(f(θ))}^{2}}dθ \\ & ={∫}_{α}^{β}\sqrt{{r}^{2}+{(\frac{dr}{dθ})}^{2}}dθ.\end{array}\]

This gives us the following theorem.

Key Concepts

• The area of a region in polar coordinates defined by the equation \(r=f(θ)\) with \(α\le θ\le β\) is given by the integral \(A=\frac{1}{2}{{∫}_{α}^{β}[f(θ)]}^{2}dθ.\) (See Example 1.)
• To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas. (See Example 2.)
• The arc length of a polar curve defined by the equation \(r=f(θ)\) with \(α\le θ\le β\) is given by the integral \(L={∫}_{α}^{β}\sqrt{{[f(θ)]}^{2}+{[{f}^{\prime }(θ)]}^{2}}dθ={∫}_{α}^{β}\sqrt{{r}^{2}+{(\frac{dr}{dθ})}^{2}}dθ.\) (See Example 3.)

Key Equations