![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-21-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-22-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] |
θ=0 θ=0](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-23-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π,
θ=0 θ=0](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-24-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-25-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-26-320.jpg)
![Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.](https://image.slidesharecdn.com/4areasinpolarcoordinates-130207095515-phpapp02/85/4-areas-in-polar-coordinates-27-320.jpg)
4 areas in polar coordinates
- 1. Areas in Polar Coordinates
- 2. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, r=f(θ) θ=b θ=a
- 3. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, r=f(θ) θ=b θ=a
- 4. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, r=f(θ) θ=b θ=a
- 5. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, r=f(θ) θ=b θ=a
- 6. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, r=f(θ) θ=b θ=a
- 7. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a
- 8. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a (The Polar Area Formula) The area swept out by r = f(θ) from θ = a to θ = b is b b 1 1 A= 2 ∫r 2 dθ or 2 ∫ f(θ)2 dθ θ=a θ=a
- 9. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a (The Polar Area Formula) The area swept out by r = f(θ) from θ = a to θ = b is b b 1 1 A= 2 ∫r 2 dθ or 2 ∫ f(θ)2 dθ θ=a θ=a Example A. Draw and find the area A swept by r = f(θ) = k with 0 ≤ θ ≤ 2π.
- 10. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a (The Polar Area Formula) The area swept out by r = f(θ) from θ = a to θ = b is b b 1 1 A= 2 ∫r 2 dθ or 2 ∫ f(θ)2 dθ θ=a θ=a Example A. r = f(θ) = k Draw and find the area A swept by r = f(θ) = k with 0 ≤ θ ≤ 2π. k
- 11. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a (The Polar Area Formula) The area swept out by r = f(θ) from θ = a to θ = b is b b 1 1 A= 2 ∫r 2 dθ or 2 ∫ f(θ)2 dθ θ=a θ=a Example A. r = f(θ) = k Draw and find the area A swept by r = f(θ) = k with 0 ≤ θ ≤ 2π. k b 1 A =2 ∫ r2 dθ = θ=a
- 12. Areas in Polar Coordinates Given a polar function r = f(θ) with a ≤ θ ≤ b, it "sweeps" out an area r=f(θ) between them. θ=b θ=a (The Polar Area Formula) The area swept out by r = f(θ) from θ = a to θ = b is b b 1 1 A= 2 ∫r 2 dθ or 2 ∫ f(θ)2 dθ θ=a θ=a Example A. r = f(θ) = k Draw and find the area A swept by r = f(θ) = k with 0 ≤ θ ≤ 2π. k b 2π 2π 1 1 k2θ = π k2 A =2 ∫a r dθ = 2 2 ∫ k2 dθ = 2 θ= θ=0 θ=0
- 13. Areas in Polar Coordinates We have the following formulas cos2(θ) = 1 + cos(2θ) that express the squares of the 2 sine and cosine in terms of the sin2(θ) = 1 – cos(2θ) cosine double-angle. 2
- 14. Areas in Polar Coordinates We have the following formulas cos2(θ) = 1 + cos(2θ) that express the squares of the 2 sine and cosine in terms of the sin2(θ) = 1 – cos(2θ) cosine double-angle. 2 We simplify the algebra below c2(θ) = 1 + c(2θ) by writing cos(θ) as c(θ) 2 s (θ) = 2 1 – c(2θ) and sin(θ) as s(θ). 2
- 15. Areas in Polar Coordinates We have the following formulas cos2(θ) = 1 + cos(2θ) that express the squares of the 2 sine and cosine in terms of the sin2(θ) = 1 – cos(2θ) cosine double-angle. 2 We simplify the algebra below c2(θ) = 1 + c(2θ) by writing cos(θ) as c(θ) 2 s (θ) = 2 1 – c(2θ) and sin(θ) as s(θ). 2 We use these formulas to integrate c2(θ) and s2(θ) that shows up in f2(θ), the integrand in the polar area formula. Hence to integrate ∫ c2(θ) dθ =
- 16. Areas in Polar Coordinates We have the following formulas cos2(θ) = 1 + cos(2θ) that express the squares of the 2 sine and cosine in terms of the sin2(θ) = 1 – cos(2θ) cosine double-angle. 2 We simplify the algebra below c2(θ) = 1 + c(2θ) by writing cos(θ) as c(θ) 2 s (θ) = 2 1 – c(2θ) and sin(θ) as s(θ). 2 We use these formulas to integrate c2(θ) and s2(θ) that shows up in f2(θ), the integrand in the polar area formula. Hence to integrate ∫ c2(θ) dθ = ∫ 1 + c(2θ) dθ = 2
- 17. Areas in Polar Coordinates We have the following formulas cos2(θ) = 1 + cos(2θ) that express the squares of the 2 sine and cosine in terms of the sin2(θ) = 1 – cos(2θ) cosine double-angle. 2 We simplify the algebra below c2(θ) = 1 + c(2θ) by writing cos(θ) as c(θ) 2 s (θ) = 2 1 – c(2θ) and sin(θ) as s(θ). 2 We use these formulas to integrate c2(θ) and s2(θ) that shows up in f2(θ), the integrand in the polar area formula. Hence to integrate ∫ c2(θ) dθ = ∫ 1 + c(2θ) dθ = ½ ∫1 + c(2θ) dθ 2 = ½ (θ + ½ s(2θ)) + C or ½ θ + ¼ s(2θ) + C
- 18. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
- 19. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π.
- 20. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) 1
- 21. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
- 22. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k
- 23. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k The area swept over by r as θ goes from 0 to 2π is 2π 2π ½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | θ=0 θ=0
- 24. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k The area swept over by r as θ goes from 0 to 2π is 2π 2π ½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, θ=0 θ=0
- 25. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k The area swept over by r as θ goes from 0 to 2π is 2π 2π ½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is θ=0 θ=0 the area of a circle with radius 1, swept over twice.
- 26. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k The area swept over by r as θ goes from 0 to 2π is 2π 2π ½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is θ=0 θ=0 the area of a circle with radius 1, swept over twice.
- 27. Areas in Polar Coordinates We summarize the above integrals here. ∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C Draw and find the area swept by r = f(θ) = 2sin(θ) with θ goes from 0 to 2π. r = f(θ) = 2sin(θ) The anti–derivative ∫ f 2(θ) dθ 1 =∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ = 4(½ θ – ¼ s(2θ)) + k = 2θ – s(2θ) + k The area swept over by r as θ goes from 0 to 2π is 2π 2π ½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is θ=0 θ=0 the area of a circle with radius 1, swept over twice.
- 28. Areas in Polar Coordinates Example C. Draw and find the area enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant.
- 29. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown.
- 30. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown. It’s the swept area as π θ goes from 0 to 2 which is
- 31. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown. It’s the swept area as θ goes from 0 to 2 π
- 32. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown. It’s the swept area as π θ goes from 0 to 2 which is 1 π /2 1 π /2 2 ∫0 (1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ 2 2 0 1 1 1 π /2 = (θ − 2 sin(θ ) + sin( 2θ ) + θ ) | 2 4 2 0 1 3 1 π /2 1 3 π 3π = ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1 2 2 4 0 2 2 2 8
- 33. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown. It’s the swept area as π θ goes from 0 to 2 which is 1 π /2 1 π /2 2 ∫0 (1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ 2 2 0 1 1 1 π /2 = (θ − 2 sin(θ ) + sin( 2θ ) + θ ) | 2 4 2 0 1 3 1 π /2 1 3 π 3π = ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1 2 2 4 0 2 2 2 8
- 34. Areas in Polar Coordinates Example C. Draw and find the area θ=π/2 enclosed by r = f(θ) = 1 – cos(θ) in the first quadrant. 2 The region in question is θ=0 shown. It’s the swept area as π θ goes from 0 to 2 which is 1 π /2 1 π /2 2 ∫0 (1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ 2 2 0 1 1 1 π /2 = (θ − 2 sin(θ ) + sin( 2θ ) + θ ) | 2 4 2 0 1 3 1 π /2 1 3 π 3π = ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1 2 2 4 0 2 2 2 8
- 35. Areas in Polar Coordinates Given polar functions R = f(θ) outer R = f(θ) and r = g(θ) where R > r for θ between θ=a a and b, θ=b inner r = g(θ)
- 36. Areas in Polar Coordinates Given polar functions R = f(θ) outer R = f(θ) and r = g(θ) where R > r for θ between θ=a a and b, then the area θ=b inner r = g(θ) between them is b b ½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ θ=a θ=a
- 37. Areas in Polar Coordinates Given polar functions R = f(θ) outer R = f(θ) and r = g(θ) where R > r for θ between θ=a a and b, then the area θ=b inner r = g(θ) between them is b b ½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ θ=a θ=a To find the area enclosed by two polar graphs, the first step is to determine the angles θ = a and θ = b that wedge in the area.
- 38. Areas in Polar Coordinates Given polar functions R = f(θ) outer R = f(θ) and r = g(θ) where R > r for θ between θ=a a and b, then the area θ=b inner r = g(θ) between them is b b ½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ θ=a θ=a To find the area enclosed by two polar graphs, the first step is to determine the angles θ = a and θ = b that wedge in the area. This should be done both algebraically and graphically because the algebraic solution might be incomplete.
- 39. Areas in Polar Coordinates Given polar functions R = f(θ) outer R = f(θ) and r = g(θ) where R > r for θ between θ=a a and b, then the area θ=b inner r = g(θ) between them is b b ½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ θ=a θ=a To find the area enclosed by two polar graphs, the first step is to determine the angles θ = a and θ = b that wedge in the area. This should be done both algebraically and graphically because the algebraic solution might be incomplete. Next we have to split the area into separate wedges radialIy from corner to corner.
- 40. Areas in Polar Coordinates Example D. R = 2cos(θ) Find the shaded area r=1 shown in the figure.
- 41. Areas in Polar Coordinates Example D. R = 2cos(θ) Find the shaded area r=1 shown in the figure. First set the two curves equal to each other to find the angles of the intersection. 1 = 2cos(θ)
- 42. Areas in Polar Coordinates Example D. R = 2cos(θ) Find the shaded area r=1 shown in the figure. First set the two curves equal to each other to find the angles of the intersection. 1 = 2cos(θ) cos(θ)=1/2 so θ = π/3
- 43. Areas in Polar Coordinates Example D. θ=π/2 θ=π/3 R = 2cos(θ) Find the shaded area r=1 shown in the figure. θ=0 First set the two curves equal to each other to find the angles of the intersection. 1 = 2cos(θ) cos(θ)=1/2 so θ = π/3 The area consists of two regions. i. the red area with the r = 1 as boundary with θ goes from 0 to π/3.
- 44. Areas in Polar Coordinates Example D. θ=π/2 θ=π/3 R = 2cos(θ) Find the shaded area r=1 shown in the figure. θ=0 First set the two curves equal to each other to find the angles of the intersection. 1 = 2cos(θ) cos(θ)=1/2 so θ = π/3 The area consists of two regions. i. the red area with the r = 1 as boundary with θ goes from 0 to π/3. ii. the black area with R = 2cos(θ) as the boundary with θ going from π/3 to θ = π/2.
- 45. Areas in Polar Coordinates Example D. θ=π/2 θ=π/3 R = 2cos(θ) Find the shaded area r=1 shown in the figure. θ=0 First set the two curves equal to each other to find the angles of the intersection. 1 = 2cos(θ) cos(θ)=1/2 so θ = π/3 The area consists of two regions. i. the red area with the r = 1 as boundary with θ goes from 0 to π/3. ii. the black area with R = 2cos(θ) as the boundary with θ going from π/3 to θ = π/2. The answer is the sum of two areas
- 46. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0
- 47. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0 The black area are bounded by θ = π/3 to θ = π/2 is
- 48. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0 The black area are bounded by θ = π/3 to θ = π/2 is 1 π /2 2 ∫π / 3 (2 cos(θ )) 2 dθ π /2 1 1 π /2 = 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) | 2 π /3 4 2 π /3 1 π /2 π 1 2π π π 3 = sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = − 2 π /3 2 2 3 3 6 4 The total area is π + π – √3 = π – √3 6 6 4 3 4
- 49. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0 The black area are bounded by θ = π/3 to θ = π/2 is 1 π /2 2 ∫π / 3 (2 cos(θ )) 2 dθ π /2 1 1 π /2 = 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) | 2 π /3 4 2 π /3 1 π /2 π 1 2π π π 3 = sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = − 2 π /3 2 2 3 3 6 4 The total area is π + π – √3 = π – √3 6 6 4 3 4
- 50. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0 The black area are bounded by θ = π/3 to θ = π/2 is 1 π /2 2 ∫π / 3 (2 cos(θ )) 2 dθ π /2 1 1 π /2 = 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) | 2 π /3 4 2 π /3 1 π /2 π 1 2π π π 3 = sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = − 2 π /3 2 2 3 3 6 4
- 51. Areas in Polar Coordinates The red area bounded from θ=π/2 θ=π/3 r=1 R = 2cos(θ) θ = 0 to θ=π/3 is 1/6 of the unit circle. It's area is π/6. θ=0 The black area are bounded by θ = π/3 to θ = π/2 is 1 π /2 2 ∫π / 3 (2 cos(θ )) 2 dθ π /2 1 1 π /2 = 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) | 2 π /3 4 2 π /3 1 π /2 π 1 2π π π 3 = sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = − 2 π /3 2 2 3 3 6 4 The total area is π + π – √3 = π – √3 6 6 4 3 4
- 52. Areas in Polar Coordinates Method II: The blue area is θ=π/2 θ=π/3 R = 2cos(θ) ¼ of a circle minus the red r=1 area.
- 53. Areas in Polar Coordinates Method II: The blue area is θ=π/2 θ=π/3 R = 2cos(θ) ¼ of a circle minus the red r=1 area. The red area is bounded by f(θ)=1 and g(θ)=2cos(θ) where θ goes from π/3 to π/2.
- 54. Areas in Polar Coordinates Method II: The blue area is θ=π/2 θ=π/3 R = 2cos(θ) ¼ of a circle minus the red r=1 area. The red area is bounded by f(θ)=1 and g(θ)=2cos(θ) where θ goes from π/3 to π/2. It's area is (outer curve)2 (inner curve)2 1 π/2 2 2 ∫ /3 1 − (2 cos(θ )) 2 dθ π 1 π/2 2 π/2 = ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ 2 π /3 π /3 1 π π π 3 π 3 = ( − ) −( − ) =− + 2 2 3 6 4 12 4 So the blue area is π – (– π + √3 ) = π – √3 4 12 4 3 4
- 55. Areas in Polar Coordinates Method II: The blue area is θ=π/2 θ=π/3 R = 2cos(θ) ¼ of a circle minus the red r=1 area. The red area is bounded by f(θ)=1 and g(θ)=2cos(θ) where θ goes from π/3 to π/2. It's area is (outer curve)2 (inner curve)2 1 π/2 2 2 ∫ /3 1 − (2 cos(θ )) 2 dθ π 1 π/2 2 π/2 = ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ 2 π /3 π /3 1 π π π 3 π 3 = ( − ) −( − ) =− + 2 2 3 6 4 12 4
- 56. Areas in Polar Coordinates Method II: The blue area is θ=π/2 θ=π/3 R = 2cos(θ) ¼ of a circle minus the red r=1 area. The red area is bounded by f(θ)=1 and g(θ)=2cos(θ) where θ goes from π/3 to π/2. It's area is (outer curve)2 (inner curve)2 1 π/2 2 2 ∫ /3 1 − (2 cos(θ )) 2 dθ π 1 π/2 2 π/2 = ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ 2 π /3 π /3 1 π π π 3 π 3 = ( − ) −( − ) =− + 2 2 3 6 4 12 4 So the blue area is π – (– π + √3 ) = π – √3 4 12 4 3 4
- 57. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2
- 58. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded.
- 59. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area.
- 60. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area. Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer is 2θ = π/6, or that θ = π/12.
- 61. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area. Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer is 2θ = π/6, or that θ = π/12. From the symmetry of the graph the other θ is 5π/12.
- 62. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area. Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer is 2θ = π/6, or that θ = π/12. From the symmetry of the graph the other θ is 5π/12.
- 63. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area. Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer is 2θ = π/6, or that θ = π/12. From the symmetry of the graph the other θ is 5π/12. Hence one piece is (outer curve)2 (inner curve)2 1 5π /12 5π / 12 2π ∫π /12 (4 sin(2θ )) −2 dθ = ∫π /12 8 sin (2θ ) − 2dθ = 3 + 3 2 2 2 2
- 64. Areas in Polar Coordinates Example E. Find the area that R = 4sin(2θ) is outside of r = 2 and inside of R = 4sin(2θ). r=2 We only need to find one of the four equal regions as shaded. We need to find the angles of intersections that wedge in the area. Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer is 2θ = π/6, or that θ = π/12. From the symmetry of the graph the other θ is 5π/12. Hence one piece is (outer curve)2 (inner curve)2 1 5π /12 5π / 12 2π ∫π /12 (4 sin(2θ )) −2 dθ = ∫π /12 8 sin (2θ ) − 2dθ = 3 + 3 2 2 2 2 So the total area is 4*(2π/3 + √3)