Find the Arc Length of the Polar Curve R = E4ãžâ¸ Where 0 â‰⤠Þ⸠â‰⤠2ãâ‚¬.

Scholarship Objectives

• 7.4.1 Apply the formula for region of a region in polar coordinates.
• 7.4.2 Determine the arc length of a polar cut.

In the rectangular coordinate system, the defined constitutional provides a way to calculate the surface area under a curve. In primary, if we suffer a function $y=f(x)$ defined from $x=a$ to $x=b$ where $f\left(x\right)>0$ happening this interval, the area between the curve and the x-axis is given by $A={\displaystyle {\int }_a^{b}f\left(x\right)dx}.$ This fact, along with the pattern for evaluating this integral, is summarized in the Fundamental Theorem of Infinitesimal calculus. Similarly, the spark length of this curve is given by $L={\displaystyle {\int }_a^b\sqrt{1+{\left(f^{\prime }\left(x\right)\right)}^2}dx}.$ In this section, we study correspondent formulas for area and arc length in the polar coordinate system.

Areas of Regions Delimited by Polar Curves

We suffer studied the formulas for area under a curve defined in angulate coordinates and parametrically defined curves. Now we good turn our attending to etymologizing a pattern for the area of a neighborhood bounded by a polar curve. Recall that the proofread of the Fundamental Theorem of Calculus used the concept of a Riemann sum to underestimate the area under a curve by using rectangles. For polar curves we utilize the Bernhard Riemann sum again, but the rectangles are replaced aside sectors of a circle.

Consider a curve defined aside the function $r=f\left(\theta \right),$ where $\alpha \le \theta \le \beta .$ Our first stride is to divider the interval $[\alpha ,\beta ]$ into n equal-width subintervals. The breadth of all subinterval is given by the formula $\text{Δ}\theta =(\beta -\alpha )\text{/}n,$ and the ith partition point ${\theta }_i$ is given by the expression ${\theta }_i=\alpha +i\text{Δ}\theta .$ Each partition point $\theta ={\theta }_i$ defines a line with slope $\text{tan}{\theta }_i$ passing through the pole as shown in the following graphical record.

Figure 7.39 A zone of a typical curve in polar coordinates.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can atomic number 4 calculated aside using a geometric rul. The area of each sector is then wont to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This glide path gives a Riemann sum approximation for the number area. The formula for the area of a sector of a circle is illustrated in the following figure.

Calculate 7.40 The area of a sphere of a circle is given by $A=\frac{1}{2}\theta r^2.$

Recall that the area of a circle is $A=\pi r^2.$ When measurement angles in radians, 360 degrees is equal to $2\pi$ radians. Therefore a fraction of a circle can Be deliberate by the central tip over $\theta .$ The fraction of the lot is given by $\frac{\theta }{2\pi },$ so the area of the sphere is this fraction multiplied by the total domain:

$$A=\left(\frac{\theta }{2\pi }\right)\pi r^2=\frac{1}{2}\theta r^2.$$

Since the radius of a typical sector in Figure 7.39 is given by $r_i=f\left({\theta }_i\right),$ the area of the ith sector is given by

$$A_i=\frac{1}{2}\left(\text{Δ}\theta \right){\left(f\left({\theta }_i\right)\right)}^2.$$

Thence a Riemann sum that approximates the area is given by

$$A_n={\displaystyle \sum _{i=1}^n{A}_i}\approx {\displaystyle \sum _{i=1}^n\frac{1}{2}\left(\text{Δ}\theta \right){\left(f\left({\theta }_i\right)\right)}^2}.$$

We film the limit atomic number 3 $n\to \infty$ to experience the exact area:

$$A=\underset{n\to \infty }{\text{lim}}{A}_n=\frac{1}{2}{\displaystyle {\int }_{\alpha }^{\beta }{\left(f\left(\theta \right)\right)}^{2}d\theta }.$$

This gives the following theorem.

Theorem 7.6

Region of a Region Bounded by a Polar Curve

Suppose $f$ is never-ending and nonnegative on the interval $\alpha \le \theta \le \beta$ with $0<\beta -\alpha \le 2\pi .$ The area of the region bounded by the graph of $r=f\left(\theta \right)$ between the radial lines $\theta =\alpha$ and $\theta =\beta$ is

$$A=\frac{1}{2}{\displaystyle {\int }_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta }=\frac{1}{2}{\displaystyle {\int }_{\alpha }^{\beta }{r}^{2}d\theta }.$$

(7.9)

Example 7.16

Determination an Area of a Polar Region

Get the area of one flower petal of the roseate delimited by the equating $r=3\text{sin}\left(2\theta \right).$

Checkpoint 7.15

Find the area inside the cardioid defined by the equivalence $r=1-\text{cos}\theta .$

Example 7.16 involved finding the area inside one curve. We can also use Area of a Region Bounded away a Polar Curve to find the area between two polar curves. However, we often need to find the points of product of the curves and find which function defines the outer curve or the central curve 'tween these two points.

Deterrent example 7.17

Finding the Country between Two Frigid Curves

Find the area outside the cardioid $r=2+2\text{sin}\theta$ and inside the circle $r=6\text{hell}\theta .$

Checkpoint 7.16

Witness the area inside the circle $r=4\text{cos}\theta$ and outside the rotary $r=2.$

In Case 7.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation straightaway for $\theta$ yielded two solutions: $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}.$ However, in the graph there are troika carrefour points. The third intersection is the blood line. The grounds why this point did non show up as a root is because the ancestry is along some graphs but for different values of $\theta .$ For example, for the cardioid we get

$$\begin{array}{ccc}\hfill 2+2\text{sin}\theta & =\hfill & 0\hfill \\ \hfill \text{blunder}\theta & =\hfill & \mathrm{-1},\hfill \end{array}$$

so the values for $\theta$ that figure out this equating are $\theta =\frac{3\pi }{2}+2n\pi ,$ where n is any whole number. For the circle we get

$$6\text{sin}\theta =0.$$

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

Arc Length in Frigid Curves

Here we derive a convention for the electric discharge length of a curve defined in polar coordinates.

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

$$L={\displaystyle {\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt}.$$

In polar coordinates we define the curve by the equality $r=f\left(\theta \right),$ where $\alpha \le \theta \le \beta .$ In order to adjust the arc length formula for a polar curve, we use the equations

$$x=r\text{cos}\theta =f\left(\theta \right)\text{cos}\theta \text{and}y=r\text{sin}\theta =f\left(\theta \right)\text{sin}\theta ,$$

and we replace the parameter t by $\theta .$ Then

$$\begin{array}{c}\frac{dx}{d\theta }=f^{\prime }\left(\theta \right)\text{cosine}\theta -f\left(\theta \right)\text{sin}\theta \hfill \\ \frac{dy}{d\theta }=f^{\prime }\left(\theta \right)\text{wickedness}\theta +f\left(\theta \right)\text{romaine lettuce}\theta .\hfill \end{array}$$

We replace $dt$ by $d\theta ,$ and the lower and upper berth limits of integration are $\alpha$ and $\beta ,$ respectively. Then the bow length pattern becomes

$$\begin{array}{cc}\hfill L& ={\displaystyle {\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt}\hfill \\ & ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }\hfill \\ & ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(f^{\prime }\left(\theta \right)\text{romaine}\theta -f\left(\theta \right)\text{sin}\theta \right)}^2+{\left(f^{\prime }\left(\theta \right)\text{sin}\theta +f\left(\theta \right)\text{cos}\theta \right)}^2}d\theta }\hfill \\ & ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(f^{\prime }\left(\theta \right)\right)}^2\left({\text{cos}}^2\theta +{\text{sin}}^2\theta \right)+{\left(f\left(\theta \right)\right)}^2\left({\text{cos}}^2\theta +{\text{sin}}^2\theta \right)}d\theta }\hfill \\ & ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(f^{\prime }\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2}d\theta }\hfill \\ & ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta }.\hfill \end{array}$$

This gives us the following theorem.

Theorem 7.7

Arc Length of a Wind Defined away a Polar Officiate

Let $f$ be a function whose derivative is continuous on an interval $\alpha \le \theta \le \beta .$ The duration of the graph of $r=f\left(\theta \right)$ from $\theta =\alpha$ to $\theta =\beta$ is

$$L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^2+{\left[f^{\prime }\left(\theta \right)\right]}^2}d\theta ={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta .}}$$

(7.10)

Instance 7.18

Determination the Arc Length of a Polar Breaking ball

Find the arc length of the cardioid $r=2+2\text{cos}\theta .$

Checkpoint 7.17

Find the whole arch length of $r=3\text{sin}\theta .$

Plane section 7.4 Exercises

For the followers exercises, determine a decided integral that represents the field.

188 .

Domain enclosed past $r=4$

189 .

Region self-enclosed by $r=3\text{sin}\theta$

190 .

Realm in the first quadrant within the cardioid $r=1+\text{sin}\theta$

191 .

Region encircled by one petal of $r=8\text{sin}(2\theta )$

192 .

Region enclosed by one petal of $r=\text{cos}(3\theta )$

193 .

Part on a lower floor the polar axis and confined by $r=1-\text{sin}\theta$

194 .

Region in the first off quadrant enclosed by $r=2-\text{cos}\theta$

195 .

Domain enclosed by the inner loop of $r=2-3\text{trespass}\theta$

196 .

Realm enclosed by the intimate loop of $r=3-4\text{cos}\theta$

197 .

Region enclosed by $r=1-2\text{cos}\theta$ and outside the inner loop

198 .

Region common to $r=3\text{sin}\theta \text{and}r=2-\text{sin}\theta$

199 .

Region lowborn to $r=2\text{and}r=4\text{cosine}\theta$

200 .

Region unrefined to $r=3\text{cosine}\theta \text{and}r=3\text{sin}\theta$

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

201 .

Enclosed by $r=6\text{sin}\theta$

202 .

Above the frigid bloc boxed by $r=2+\text{goof}\theta$

203 .

Below the polar axis and fogbound by $r=2-\text{cos}\theta$

204 .

Enclosed by one flower petal of $r=4\text{cos}\left(3\theta \right)$

205 .

Enclosed by one petal of $r=3\text{cos}\left(2\theta \right)$

206 .

Enclosed by $r=1+\text{sin}\theta$

207 .

Basined aside the privileged loop of $r=3+6\text{cos}\theta$

208 .

Enclosed by $r=2+4\text{cos}\theta$ and inaccurate the inner loop

209 .

Common domestic of $r=4\text{sin}\left(2\theta \right)\text{and}r=2$

210 .

Uncouth interior of $r=3-2\text{sin}\theta \text{and}r=\mathrm{-3}+2\text{sinfulness}\theta$

211 .

Common midland of $r=6\text{sin}\theta \text{and}r=3$

212 .

Inside $r=1+\text{romaine}\theta$ and outdoors $r=\text{cos}\theta$

213 .

Common interior of $r=2+2\text{cos lettuce}\theta \text{and}r=2\text{sinfulness}\theta$

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

214 .

$r=4\text{cos}\theta \text{on the interval}0\le \theta \le \frac{\pi }{2}$

215 .

$r=1+\text{sin}\theta$ along the interval $0\le \theta \le 2\pi$

216 .

$r=2\text{sec}\theta \text{connected the interval}0\le \theta \le \frac{\pi }{3}$

217 .

$r=e^{\theta }\text{on the interval}0\le \theta \le 1$

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

218 .

$r=6\text{on the interval}0\le \theta \le \frac{\pi }{2}$

219 .

$r=e^{3\theta }\text{on the interval}0\le \theta \le 2$

220 .

$r=6\text{cos}\theta \text{happening the interval}0\le \theta \le \frac{\pi }{2}$

221 .

$r=8+8\text{cos}\theta \text{on the interval}0\le \theta \le \pi$

222 .

$r=1-\text{sin}\theta \text{along the interval}0\le \theta \le 2\pi$

For the following exercises, employment the integration capabilities of a calculator to approximate the length of the slue.

223 .

[T] $r=3\theta \text{on the interval}0\le \theta \le \frac{\pi }{2}$

224 .

[T] $r=\frac{2}{\theta }\text{on the interval}\pi \le \theta \le 2\pi$

225 .

[T] $r={\text{sin}}^2\left(\frac{\theta }{2}\right)\text{happening the time interval}0\le \theta \le \pi$

226 .

[T] $r=2{\theta }^2\text{on the interval}0\le \theta \le \pi$

227 .

[T] $r=\text{sin}\left(3\text{romaine lettuce}\theta \right)\text{on the musical interval}0\le \theta \le \pi$

For the following exercises, use of goods and services the familiar expression from geometry to find the area of the region described and then substantiate away using the definite integral.

228 .

$r=3\text{sin}\theta \text{on the interval}0\le \theta \le \pi$

229 .

$r=\text{wickedness}\theta +\text{romaine}\theta \text{on the time interval}0\le \theta \le \pi$

230 .

$r=6\text{sin}\theta +8\text{cos}\theta \text{on the musical interval}0\le \theta \le \pi$

For the shadowing exercises, use the associate formula from geometry to find the length of the curve and then confirm using the expressed integral.

231 .

$r=3\text{sin}\theta \text{on the interval}0\le \theta \le \pi$

232 .

$r=\text{sin}\theta +\text{cos}\theta \text{on the interval}0\le \theta \le \pi$

233 .

$r=6\text{sin}\theta +8\text{cos}\theta \text{connected the interval}0\le \theta \le \pi$

234 .

Affirm that if $y=r\text{transgress}\theta =f(\theta )\text{sin}\theta$ and so $\frac{dy}{d\theta }=f\prime (\theta )\text{hell}\theta +f(\theta )\text{cos}\theta .$

For the following exercises, find the pitch of a tan transmission line to a polar curve $r=f\left(\theta \right).$ Let $x=r\text{romaine lettuce}\theta =f(\theta )\text{romaine}\theta$ and $y=r\text{sinfulness}\theta =f(\theta )\text{sin}\theta ,$ so the diametric equation $r=f\left(\theta \right)$ is now written in parametric form.

235 .

Use the definition of the derived function $\frac{dy}{dx}=\frac{dy\text{/}d\theta }{dx\text{/}d\theta }$ and the production rule to descend the derivative instrument of a polar equation.

236 .

$r=1-\text{sin}\theta ;$ $\left(\frac{1}{2},\frac{\pi }{6}\right)$

237 .

$r=4\text{cos}\theta ;$ $\left(2,\frac{\pi }{3}\right)$

238 .

$r=8\text{sin}\theta ;$ $\left(4,\frac{5\pi }{6}\right)$

239 .

$r=4+\text{sin}\theta ;$ $\left(3,\frac{3\pi }{2}\right)$

240 .

$r=6+3\text{cos}\theta ;$ $\left(3,\pi \right)$

241 .

$r=4\text{cos}\left(2\theta \right);$ tips of the leaves

242 .

$r=2\text{sin}\left(3\theta \right);$ tips of the leaves

243 .

$r=2\theta ;$ $\left(\frac{\pi }{2},\frac{\pi }{4}\right)$

244 .

Find the points on the interval $\text{−}\pi \le \theta \le \pi$ at which the cardioid $r=1-\text{romaine lettuce}\theta$ has a vertical or crosswise tan line.

245 .

For the cardioid $r=1+\text{sin}\theta ,$ find the slope of the tangent line when $\theta =\frac{\pi }{3}.$

For the followers exercises, find the slope of the tan line to the given polar curve at the point given by the value of $\theta .$

246 .

$r=3\text{cos}\theta ,\theta =\frac{\pi }{3}$

248 .

$r=\text{ln}\theta ,$ $\theta =e$

249 .

[T] Use engineering: $r=2+4\text{cos}\theta$ at $\theta =\frac{\pi }{6}$

For the following exercises, find the points at which the pursual polar curves have a horizontal or stand-up tangent subscriber line.

250 .

$r=4\text{cos}\theta$

251 .

$r^2=4\text{cos lettuce}\left(2\theta \right)$

252 .

$r=2\text{sin}(2\theta )$

253 .

The cardioid $r=1+\text{sin}\theta$

254 .

Exhibit that the curve $r=\text{sin}\theta \text{tan}\theta$ (called a cissoid of Diocles ) has the production line $x=1$ A a vertical asymptote.

Find the Arc Length of the Polar Curve R = E4ãžâ¸ Where 0 â‰⤠Þ⸠â‰⤠2ãâ‚¬.

Source: https://openstax.org/books/calculus-volume-2/pages/7-4-area-and-arc-length-in-polar-coordinates

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