(a) Explanation Given: The 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 : V = π ∫ x 1 x 2 f 2 ( x ) d x The symmetrical property of integration: ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x The property of integral: ∫ ( a + b ) d x = ∫ a d x + ∫ b d x The formula for eccentricity of an ellipse: e = 1 − b 2 a 2 The integration formula for n th power of sec: ∫ 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. Rewrite 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) Use disk integration to 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 ) Solve 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 Simplify 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 . Now, 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 Assume 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 Put 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 Put 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 Simplify 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 Solve the above integral as below: For ellipse, Centre of ellipse c 2 = a 2 − b 2 . Therefore, 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 solid generated by revolving the region about the major axis of the ellipse has a volume of V = 4 π a b 2 3 and surface area of S = 2 π b 2 + 2 π ( a b e ) sin − 1 e .

7th Edition

ISBN: 8220106798560

Author: Edwards

Publisher: YUZU

See similar textbooks

Question

Chapter 10, Problem 8PS

(a)

To determine

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

(b)

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10, Problem 7PS

Chapter 10, Problem 9PS

Students have asked these similar questions

Why does theta go from -pi/2 to pi/2 ?

A) Use Green's theorem to compute the area inside the ellipse Use the fact that the area can be written as Hint: x (t) = 17 cos(t). The area is Jd dx dy = 1/2 SOD ૨૨ 17² -y dx + x dy. + y² 15² = 1. B) Find a parametrization of the curve 2/3 + y2/3 = 42/3 and use it to compute the area of the interior. Hint: x (t) = 4 cos³ (t).

1. Find the volume of an ellipsoid obtained by rotating the region confined between the y-axis and the graph of ellipse x2 = 1. 62 a2 (a) Sketch carefully the graph of an ellipse. Feel free to google it, but you should learn how to provide a plot of an ellipse given arbitrary positive constants a and b. (b) Identify the region that you will rotate around the y-axis and show it on your sketch. (c) Use the method of disks to find the volume (set up the integral and evaluate it providing all the details).. (d) Use the method of cylindrical shells to find the volume (set up the integral and evaluate it providing all the details). (e) Did you get the same answers in the two previous points? (f) Make a correct guess without any calculations about the volume of an ellipsoid obtained by rotating around the x-axis the region between the x-axis and the graph of the same ellipse.

Chapter 10 Solutions

EBK CALCULUS: EARLY TRANSCENDENTAL FUNC

Ch. 10.1 - Conic Sections State the definitions of parabola,...Ch. 10.1 - Reflective Property Use a sketch to illustrate the...Ch. 10.1 - Eccentricity Consider an ellipse with eccentricity...Ch. 10.1 - Prob. 4E Ch. 10.1 - Match the following graph with its equations y2=4x...Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

Knowledge Booster

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

Solution Summary: The author explains that the solid generated by revolving the region about the minor axis of the ellipse has a volume of V=4pi a2

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

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

Want to see the full answer?

Chapter 10 Solutions