(a) Explanation Given: The provided equation of ellipse: x 2 a 2 + y 2 b 2 = 1 The eccentricity of the ellipse is: e = c a Formula used: The disk integration is given as: V = π ∫ x 1 x 2 f 2 ( x ) d x The symmetrical property of integration is as given below: ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x The property of integral is as given below: ∫ ( a + b ) d x = ∫ a d x + ∫ b d x The formula for eccentricity of an ellipse is: e = 1 − b 2 a 2 The integration formula for n th power of secant function is: ∫ sec n θ d θ = n − 2 n − 1 ∫ sec n − 2 θ d θ + sec n − 2 θ tan θ n − 1 Proof: The equation of ellipse is: x 2 a 2 + y 2 b 2 = 1 Assume b be the minor axis. Rearrange the ellipse equation as below: x 2 a 2 + y 2 b 2 = 1 x 2 a 2 = 1 − y 2 b 2 x 2 = a 2 ( b 2 − y 2 b 2 ) . x 2 = a 2 b 2 ( b 2 − y 2 ) …...…... (1) Using disk integration, find the volume: π ∫ − b b x 2 d y = 2 π ∫ 0 b a 2 b 2 ( b 2 − y 2 ) d y = 2 π a 2 b 2 ∫ 0 b ( b 2 − y 2 ) d y = 2 π a 2 b 2 ( ∫ 0 b b 2 d y − ∫ 0 b y 2 d y ) Integrate the integrals as follows: 2 π a 2 b 2 ( ∫ 0 b b 2 d y − ∫ 0 b y 2 d y ) = 2 π a 2 b 2 ( b 2 [ y ] 0 b − [ y 3 3 ] 0 b ) = 2 π a 2 b 2 ( b 2 × b − 1 3 ( b 3 ) ) = 2 π a 2 b 2 ( 3 b 3 − b 3 3 ) = 2 π a 2 b 2 × 2 b 3 3 On simplifying the above equation: 2 π a 2 b 2 × 2 b 3 3 = 4 3 π a 2 b Hence, the volume of the ellipse is 4 3 π a 2 b . Take square root on both the sides of equation (1): x 2 = a 2 b 2 ( b 2 − y 2 ) x = a b b 2 − y 2 Differentiate x = a b b 2 − y 2 with respect to y : d x d y = d d y ( a b b 2 − y 2 ) = a b d d y ( b 2 − y 2 ) = a b × 1 2 b 2 − y 2 ( − 2 y ) = − a y b b 2 − y 2 Let d s be the small element of the surface of an ellipse, which can be written as: d s = 1 + ( d x d y ) 2 d y Substitute the values of d x d y = − a y b b 2 − y 2 in the above equation: 1 + ( d x d y ) 2 d y = 1 + ( − a y b b 2 − y 2 ) 2 d y = 1 + a 2 y 2 b 2 ( b 2 − y 2 ) d y = b 2 ( b 2 − y 2 ) + a 2 y 2 b 2 ( b 2 − y 2 ) d y = b 4 − b 2 y 2 + a 2 y 2 b b 2 − y 2 d y Now, the surface area along y − axis is: s = 4 π ∫ 0 b x d s Substitute the values x = a b b 2 − y 2 and d s = b 4 − b 2 y 2 + a 2 y 2 b b 2 − y 2 d y in the above equation: s = 4 π ∫ 0 b x d s = 4 π ∫ 0 b a b b 2 − y 2 × b 4 − b 2 y 2 + a 2 y 2 b b 2 − y 2 d y = 4 π ∫ 0 b a b 2 b 4 − b 2 y 2 + a 2 y 2 d y = 4 π a b 2 ∫ 0 b ( b 4 − b 2 y 2 + a 2 y 2 ) d y On simplifying the above equation: 4 π a b 2 ∫ 0 b ( b 4 − b 2 y 2 + a 2 y 2 ) d y = 4 π a b 2 ∫ 0 b ( b 4 + y 2 ( a 2 − b 2 ) ) d y Simplify the above integral as below: For ellipse, Centre of ellipse c 2 = a 2 − b 2 . So, 4 π a b 2 ∫ 0 b ( b 4 + y 2 ( a 2 − b 2 ) ) d y = 4 π a b 2 ∫ 0 b ( b 4 + y 2 c 2 ) d y …...….. (b) To determine To prove: The volume of the solid generated by revolving the region about the major axis of the ellipse is V = 4 π a b 2 3 and surface area of the same solid is S = 2 π b 2 + 2 π ( a b e ) sin − 1 e .

11th Edition

ISBN: 9781337275361

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 10, Problem 8PS

(a)

To determine

To prove: The volume of a solid generated by revolving the region about the minor axis is V=4πa2b3 and surface area for the same solid is S=2πa2+π(b2e)ln(1+e1−e).

(b)

To determine

To prove: The volume of the solid generated by revolving the region about the major axis of the ellipse is V=4πab23 and surface area of the same solid is S=2πb2+2π(abe)sin−1e.

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Chapter 10, Problem 7PS

Chapter 10, Problem 9PS

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To prove: The volume of a solid generated by revolving the region about the minor axis is V = 4 π a 2 b 3 and surface area for the same solid is S = 2 π a 2 + π ( b 2 e ) ln ( 1 + e 1 − e ) .

Solution Summary: The author explains that the volume of a solid generated by revolving the region about the minor axis is V=4pi a2b3

To prove: The volume of a solid generated by revolving the region about the minor axis is V=4πa2b3 and surface area for the same solid is S=2πa2+π(b2e)ln(1+e1−e).

To prove: The volume of the solid generated by revolving the region about the major axis of the ellipse is V=4πab23 and surface area of the same solid is S=2πb2+2π(abe)sin−1e.

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Chapter 10 Solutions