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Conics, Parametric Equations, and Polar Coordinates

10.5 Area and Arc Length in Polar Coordinates

Focus on polar area, polar arc length, check tracing bounds.

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Ideas in this lesson

01Polar areaWorked steps · five questions02Polar arc lengthWorked steps · five questions03Check tracing boundsWorked steps · five questions
Idea 1 of 3

Polar area

Integrate one half of r2r^{2} over angle bounds that trace the desired region once.

Worked example 1

Find the area enclosed by r=2r = 2 divided by π\pi .
  1. A=12∫r2dθA = \dfrac{1}{2}\int r^{2} d\theta
  2. A=12∫02π4 dθA = \dfrac{1}{2}\int_{0}^{2\pi } 4\,d\theta
  3. Aπ=4\dfrac{A}{\pi } = 4

Answer: 44

Worked example 2

Find the area enclosed by r=3cos⁡θr = 3 \cos \theta divided by π\pi .
  1. A=12∫r2dθA = \dfrac{1}{2}\int r^{2} d\theta
  2. A=12∫−π2π29cos⁡2θ dθA = \dfrac{1}{2}\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}} 9 \cos ^{2}\theta \,d\theta
  3. Aπ=94≈2.25\dfrac{A}{\pi } = \dfrac{9}{4} \approx 2.25

Answer: 94≈2.25\dfrac{9}{4} \approx 2.25

Idea 2 of 3

Polar arc length

Use ∫ r2+(drdθ)2dθ\int\,\sqrt{r^{2} + (\frac{dr}{d}\theta )^{2}} d\theta . The angular variable must be measured in radians.

Worked example 1

For r=2r = 2 over 0≤θ≤1π20 \le \theta \le \dfrac{1\pi }{2}, find the arc length divided by π\pi .
  1. drdθ=0\dfrac{dr}{d}\theta = 0; r2+(drdθ)2=2\sqrt{r^{2} + (\dfrac{dr}{d}\theta )^{2}} = 2
  2. Lπ=2π∫01π21 dθ=1\dfrac{L}{\pi } = \dfrac{2}{\pi }\int_{0}^{\dfrac{1\pi }{2}} 1\,d\theta = 1

Answer: 11

Worked example 2

For r=3r = 3 over 0≤θ≤2π20 \le \theta \le \dfrac{2\pi }{2}, find the arc length divided by π\pi .
  1. drdθ=0\dfrac{dr}{d}\theta = 0; r2+(drdθ)2=3\sqrt{r^{2} + (\dfrac{dr}{d}\theta )^{2}} = 3
  2. Lπ=3π∫02π21 dθ=3\dfrac{L}{\pi } = \dfrac{3}{\pi }\int_{0}^{\dfrac{2\pi }{2}} 1\,d\theta = 3

Answer: 33

Idea 3 of 3

Check tracing bounds

A polar curve can be traced more than once over a long interval. Determine the actual tracing interval before computing area.

Worked example 1

Find the area enclosed by r=2r = 2 divided by π\pi .
  1. A=12∫r2dθA = \dfrac{1}{2}\int r^{2} d\theta
  2. A=12∫02π4 dθA = \dfrac{1}{2}\int_{0}^{2\pi } 4\,d\theta
  3. Aπ=4\dfrac{A}{\pi } = 4

Answer: 44

Worked example 2

Find the area enclosed by r=3cos⁡θr = 3 \cos \theta divided by π\pi .
  1. A=12∫r2dθA = \dfrac{1}{2}\int r^{2} d\theta
  2. A=12∫−π2π29cos⁡2θ dθA = \dfrac{1}{2}\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}} 9 \cos ^{2}\theta \,d\theta
  3. Aπ=94≈2.25\dfrac{A}{\pi } = \dfrac{9}{4} \approx 2.25

Answer: 94≈2.25\dfrac{9}{4} \approx 2.25

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §10.5,p10.5, p. 725725. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.