Explanation Given: The equation of the center of the circle is x 2 + y 2 = r 2 . Calculation: Show the equation as below: x 2 + y 2 = r 2 (1) The equation of line OP is shown below: y = x tan θ (2) Substitute x tan θ for y in equation (1). x 2 + ( x tan θ ) 2 = r 2 x 2 + x 2 tan 2 θ = r 2 x 2 ( 1 + tan 2 θ ) = r 2 x 2 sec 2 θ = r 2 x 2 = r 2 sec 2 θ x 2 = r 2 cos 2 θ x = r cos θ Substitute r cos θ for x in equation (2). y = r cos θ tan θ = r sin θ Calculate the area ( A 1 ) of Δ P O Q : A 1 = 1 2 b h (3) Substitute ( r cos θ ) for b and ( r sin θ ) for h in Equation (3). A 1 = 1 2 × r cos θ × r sin θ = 1 2 r 2 cos θ sin θ (4) Calculate the area ( A 2 ) of region P Q R : A 2 = ∫ a b y d x (5) Rearrange Equation (1). y 2 = r 2 − x 2 y = r 2 − x 2 Substitute r for a , r cos θ for b , and ( r 2 − x 2 ) for y in Equation (5). A 2 = ∫ r cos θ r r 2 − x 2 d x (6) Consider x = r cos u (7) Differentiate both sides of the Equation (7). d x = − r sin u d u Substitute ( r cos u ) for x and ( − r sin u d u ) for dx in Equation (6). A 2 = ∫ r cos θ r r 2 − ( r cos u ) 2 ( − r sin u d u ) = − r ∫ r cos θ r r 2 − r 2 cos 2 u sin u d u = − r ∫ r cos θ r r 2 ( 1 − cos 2 u ) sin u d u = − r ∫ 0 π 2 r sin 2 u sin u d u = − r ∫ r cos θ r r sin u sin u d u = − r 2 ∫ r cos θ r sin 2 u d u = − r 2 ∫ r cos θ r 1 2 ( 1 − cos 2 u ) d u = − r 2 2 ∫ r cos θ r ( 1 − cos 2 u ) d θ (8) Integrate Equation (8)

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Chapter 6.2, Problem 69E

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To prove: The formula (A=12r2θ) of area of a sector of a circle with radius r and central angle θ.

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Chapter 6.2, Problem 68E

Chapter 6.2, Problem 70E

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Chapter 6 Solutions

Essential Calculus: Early Transcendentals

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To prove: The formula ( A = 1 2 r 2 θ ) of area of a sector of a circle with radius r and central angle θ .

Solution Summary: The author explains how the equation of the center of a circle is x2+y2.

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To prove: The formula (A=12r2θ) of area of a sector of a circle with radius r and central angle θ.

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Chapter 6 Solutions