Calculus Volume 3

# 1.4Area and Arc Length in Polar Coordinates

Calculus Volume 31.4 Area and Arc Length in Polar Coordinates

### Learning Objectives

• 1.4.1 Apply the formula for area of a region in polar coordinates.
• 1.4.2 Determine the arc length of a polar curve.

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 $y=f(x)y=f(x)$ defined from $x=ax=a$ to $x=bx=b$ where $f(x)>0f(x)>0$ on this interval, the area between the curve and the x-axis is given by $A=∫abf(x)dx.A=∫abf(x)dx.$ 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 $L=∫ab1+(f′(x))2dx.L=∫ab1+(f′(x))2dx.$ 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 $r=f(θ),r=f(θ),$ where $α≤θ≤β.α≤θ≤β.$ Our first step is to partition the interval $[α,β][α,β]$ into n equal-width subintervals. The width of each subinterval is given by the formula $Δθ=(β−α)/n,Δθ=(β−α)/n,$ and the ith partition point $θiθi$ is given by the formula $θi=α+iΔθ.θi=α+iΔθ.$ Each partition point $θ=θiθ=θi$ defines a line with slope $tanθitanθi$ passing through the pole as shown in the following graph.

Figure 1.39 A partition of a typical curve in polar coordinates.

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.

Figure 1.40 The area of a sector of a circle is given by $A=12θr2.A=12θr2.$

Recall that the area of a circle is $A=πr2.A=πr2.$ When measuring angles in radians, 360 degrees is equal to $2π2π$ radians. Therefore a fraction of a circle can be measured by the central angle $θ.θ.$ The fraction of the circle is given by $θ2π,θ2π,$ so the area of the sector is this fraction multiplied by the total area:

$A=(θ2π)πr2=12θr2.A=(θ2π)πr2=12θr2.$

Since the radius of a typical sector in Figure 1.39 is given by $ri=f(θi),ri=f(θi),$ the area of the ith sector is given by

$Ai=12(Δθ)(f(θi))2.Ai=12(Δθ)(f(θi))2.$

Therefore a Riemann sum that approximates the area is given by

$An=∑i=1nAi≈∑i=1n12(Δθ)(f(θi))2.An=∑i=1nAi≈∑i=1n12(Δθ)(f(θi))2.$

We take the limit as $n→∞n→∞$ to get the exact area:

$A=limn→∞An=12∫αβ(f(θ))2dθ.A=limn→∞An=12∫αβ(f(θ))2dθ.$

This gives the following theorem.

Theorem 1.6

#### Area of a Region Bounded by a Polar Curve

Suppose $ff$ is continuous and nonnegative on the interval $α≤θ≤βα≤θ≤β$ with $0<β−α≤2π.0<β−α≤2π.$ The area of the region bounded by the graph of $r=f(θ)r=f(θ)$ between the radial lines $θ=αθ=α$ and $θ=βθ=β$ is

$A=12∫αβ[f(θ)]2dθ=12∫αβr2dθ.A=12∫αβ[f(θ)]2dθ=12∫αβr2dθ.$
1.9

### Example 1.16

#### Finding an Area of a Polar Region

Find the area of one petal of the rose defined by the equation $r=3sin(2θ).r=3sin(2θ).$

Checkpoint 1.15

Find the area inside the cardioid defined by the equation $r=1−cosθ.r=1−cosθ.$

Example 1.16 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 1.17

#### Finding the Area between Two Polar Curves

Find the area outside the cardioid $r=2+2sinθr=2+2sinθ$ and inside the circle $r=6sinθ.r=6sinθ.$

Checkpoint 1.16

Find the area inside the circle $r=4cosθr=4cosθ$ and outside the circle $r=2.r=2.$

In Example 1.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $θθ$ yielded two solutions: $θ=π6θ=π6$ and $θ=5π6.θ=5π6.$ 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 $θ.θ.$ For example, for the cardioid we get

$2+2sinθ=0sinθ=−1,2+2sinθ=0sinθ=−1,$

so the values for $θθ$ that solve this equation are $θ=3π2+2nπ,θ=3π2+2nπ,$ where n is any integer. For the circle we get

$6sinθ=0.6sinθ=0.$

The solutions to this equation are of the form $θ=nπθ=nπ$ 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 $(x(t),y(t))(x(t),y(t))$ for $a≤t≤ba≤t≤b$ is given by

$L=∫ab(dxdt)2+(dydt)2dt.L=∫ab(dxdt)2+(dydt)2dt.$

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

$x=rcosθ=f(θ)cosθandy=rsinθ=f(θ)sinθ,x=rcosθ=f(θ)cosθandy=rsinθ=f(θ)sinθ,$

and we replace the parameter t by $θ.θ.$ Then

$dxdθ=f′(θ)cosθ−f(θ)sinθdydθ=f′(θ)sinθ+f(θ)cosθ.dxdθ=f′(θ)cosθ−f(θ)sinθdydθ=f′(θ)sinθ+f(θ)cosθ.$

We replace $dtdt$ by $dθ,dθ,$ and the lower and upper limits of integration are $αα$ and $β,β,$ respectively. Then the arc length formula becomes

$L=∫ab(dxdt)2+(dydt)2dt=∫αβ(dxdθ)2+(dydθ)2dθ=∫αβ(f′(θ)cosθ−f(θ)sinθ)2+(f′(θ)sinθ+f(θ)cosθ)2dθ=∫αβ(f′(θ))2(cos2θ+sin2θ)+(f(θ))2(cos2θ+sin2θ)dθ=∫αβ(f′(θ))2+(f(θ))2dθ=∫αβr2+(drdθ)2dθ.L=∫ab(dxdt)2+(dydt)2dt=∫αβ(dxdθ)2+(dydθ)2dθ=∫αβ(f′(θ)cosθ−f(θ)sinθ)2+(f′(θ)sinθ+f(θ)cosθ)2dθ=∫αβ(f′(θ))2(cos2θ+sin2θ)+(f(θ))2(cos2θ+sin2θ)dθ=∫αβ(f′(θ))2+(f(θ))2dθ=∫αβr2+(drdθ)2dθ.$

This gives us the following theorem.

Theorem 1.7

#### Arc Length of a Curve Defined by a Polar Function

Let $ff$ be a function whose derivative is continuous on an interval $α≤θ≤β.α≤θ≤β.$ The length of the graph of $r=f(θ)r=f(θ)$ from $θ=αθ=α$ to $θ=βθ=β$ is

$L=∫αβ[f(θ)]2+[f′(θ)]2dθ=∫αβr2+(drdθ)2dθ.L=∫αβ[f(θ)]2+[f′(θ)]2dθ=∫αβr2+(drdθ)2dθ.$
1.10

### Example 1.18

#### Finding the Arc Length of a Polar Curve

Find the arc length of the cardioid $r=2+2cosθ. r=2+2cosθ.$

Checkpoint 1.17

Find the total arc length of $r=3sinθ.r=3sinθ.$

### Section 1.4 Exercises

For the following exercises, determine a definite integral that represents the area.

188.

Region enclosed by $r=4r=4$

189.

Region enclosed by $r=3sinθr=3sinθ$

190.

Region in the first quadrant within the cardioid $r=1+sinθr=1+sinθ$

191.

Region enclosed by one petal of $r=8sin(2θ)r=8sin(2θ)$

192.

Region enclosed by one petal of $r=cos(3θ)r=cos(3θ)$

193.

Region below the polar axis and enclosed by $r=1−sinθr=1−sinθ$

194.

Region in the first quadrant enclosed by $r=2−cosθr=2−cosθ$

195.

Region enclosed by the inner loop of $r=2−3sinθr=2−3sinθ$

196.

Region enclosed by the inner loop of $r=3−4cosθr=3−4cosθ$

197.

Region enclosed by $r=1−2cosθr=1−2cosθ$ and outside the inner loop

198.

Region common to $r=3sinθandr=2−sinθr=3sinθandr=2−sinθ$

199.

Region common to $r=2andr=4cosθr=2andr=4cosθ$

200.

Region common to $r=3cosθandr=3sinθr=3cosθandr=3sinθ$

For the following exercises, find the area of the described region.

201.

Enclosed by $r=6sinθr=6sinθ$

202.

Above the polar axis enclosed by $r=2+sinθr=2+sinθ$

203.

Below the polar axis and enclosed by $r=2−cosθr=2−cosθ$

204.

Enclosed by one petal of $r=4cos(3θ)r=4cos(3θ)$

205.

Enclosed by one petal of $r=3cos(2θ)r=3cos(2θ)$

206.

Enclosed by $r=1+sinθr=1+sinθ$

207.

Enclosed by the inner loop of $r=3+6cosθr=3+6cosθ$

208.

Enclosed by $r=2+4cosθr=2+4cosθ$ and outside the inner loop

209.

Common interior of $r=4sin(2θ)andr=2r=4sin(2θ)andr=2$

210.

Common interior of $r=3−2sinθandr=−3+2sinθr=3−2sinθandr=−3+2sinθ$

211.

Common interior of $r=6sinθandr=3r=6sinθandr=3$

212.

Inside $r=1+cosθr=1+cosθ$ and outside $r=cosθr=cosθ$

213.

Common interior of $r=2+2cosθandr=2sinθr=2+2cosθandr=2sinθ$

For the following exercises, find a definite integral that represents the arc length.

214.

$r=4cosθon the interval0≤θ≤π2r=4cosθon the interval0≤θ≤π2$

215.

$r=1+sinθr=1+sinθ$ on the interval $0≤θ≤2π0≤θ≤2π$

216.

$r=2secθon the interval0≤θ≤π3r=2secθon the interval0≤θ≤π3$

217.

$r=eθon the interval0≤θ≤1r=eθon the interval0≤θ≤1$

For the following exercises, find the length of the curve over the given interval.

218.

$r=6on the interval0≤θ≤π2r=6on the interval0≤θ≤π2$

219.

$r=e3θon the interval0≤θ≤2r=e3θon the interval0≤θ≤2$

220.

$r=6cosθon the interval0≤θ≤π2r=6cosθon the interval0≤θ≤π2$

221.

$r=8+8cosθon the interval0≤θ≤πr=8+8cosθon the interval0≤θ≤π$

222.

$r=1−sinθon the interval0≤θ≤2πr=1−sinθon the interval0≤θ≤2π$

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

223.

[T] $r=3θon the interval0≤θ≤π2r=3θon the interval0≤θ≤π2$

224.

[T] $r=2θon the intervalπ≤θ≤2πr=2θon the intervalπ≤θ≤2π$

225.

[T] $r=sin2(θ2)on the interval0≤θ≤πr=sin2(θ2)on the interval0≤θ≤π$

226.

[T] $r=2θ2on the interval0≤θ≤πr=2θ2on the interval0≤θ≤π$

227.

[T] $r=sin(3cosθ)on the interval0≤θ≤πr=sin(3cosθ)on the interval0≤θ≤π$

For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

228.

$r=3sinθon the interval0≤θ≤πr=3sinθon the interval0≤θ≤π$

229.

$r=sinθ+cosθon the interval0≤θ≤πr=sinθ+cosθon the interval0≤θ≤π$

230.

$r=6sinθ+8cosθon the interval0≤θ≤πr=6sinθ+8cosθon the interval0≤θ≤π$

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

231.

$r=3sinθon the interval0≤θ≤πr=3sinθon the interval0≤θ≤π$

232.

$r=sinθ+cosθon the interval0≤θ≤πr=sinθ+cosθon the interval0≤θ≤π$

233.

$r=6sinθ+8cosθon the interval0≤θ≤πr=6sinθ+8cosθon the interval0≤θ≤π$

234.

Verify that if $y=rsinθ=f(θ)sinθy=rsinθ=f(θ)sinθ$ then $dydθ=f′(θ)sinθ+f(θ)cosθ.dydθ=f′(θ)sinθ+f(θ)cosθ.$

For the following exercises, find the slope of a tangent line to a polar curve $r=f(θ).r=f(θ).$ Let $x=rcosθ=f(θ)cosθx=rcosθ=f(θ)cosθ$ and $y=rsinθ=f(θ)sinθ,y=rsinθ=f(θ)sinθ,$ so the polar equation $r=f(θ)r=f(θ)$ is now written in parametric form.

235.

Use the definition of the derivative $dydx=dy/dθdx/dθdydx=dy/dθdx/dθ$ and the product rule to derive the derivative of a polar equation.

236.

$r=1−sinθ;r=1−sinθ;$ $(12,π6)(12,π6)$

237.

$r=4cosθ;r=4cosθ;$ $(2,π3)(2,π3)$

238.

$r=8sinθ;r=8sinθ;$ $(4,5π6)(4,5π6)$

239.

$r=4+sinθ;r=4+sinθ;$ $(3,3π2)(3,3π2)$

240.

$r=6+3cosθ;r=6+3cosθ;$ $(3,π)(3,π)$

241.

$r=4cos(2θ);r=4cos(2θ);$ tips of the leaves

242.

$r=2sin(3θ);r=2sin(3θ);$ tips of the leaves

243.

$r=2θ;r=2θ;$ $(π2,π4)(π2,π4)$

244.

Find the points on the interval $−π≤θ≤π−π≤θ≤π$ at which the cardioid $r=1−cosθr=1−cosθ$ has a vertical or horizontal tangent line.

245.

For the cardioid $r=1+sinθ,r=1+sinθ,$ find the slope of the tangent line when $θ=π3.θ=π3.$

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of $θ.θ.$

246.

$r=3cosθ,θ=π3r=3cosθ,θ=π3$

247.

$r=θ,r=θ,$ $θ=π2θ=π2$

248.

$r=lnθ,r=lnθ,$ $θ=eθ=e$

249.

[T] Use technology: $r=2+4cosθr=2+4cosθ$ at $θ=π6θ=π6$

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

250.

$r=4cosθr=4cosθ$

251.

$r2=4cos(2θ)r2=4cos(2θ)$

252.

$r=2sin(2θ)r=2sin(2θ)$

253.

The cardioid $r=1+sinθr=1+sinθ$

254.

Show that the curve $r=sinθtanθr=sinθtanθ$ (called a cissoid of Diocles) has the line $x=1x=1$ as a vertical asymptote.