Explanation Given: The polar curves are given: r = 2 + sin 2 θ and r = 2 + cos 2 θ . Formula Used: The area of a polar curve defined by f 2 ( θ ) and f 1 ( θ ) will be described as: A = 1 2 ∫ α β ( f 2 ( θ ) 2 − f 1 ( θ ) 2 ) d θ Calculation: Consider the curve below: In the above curve, the shaded area A 1 is the area inside the curve r = 2 + sin 2 θ minus the common area between the two given curves. Now, consider the curve below: In the above curve, the shaded area A 2 is the area inside the curve r = 2 + cos 2 θ minus the common area between the two given curves. Therefore, the required area will be equal to the sum of the areas of the two given curves that is A 1 + A 2 minus two times of the common area say A 3 between the two curves. The intersection of the two given curves will be given as: 2 + cos 2 θ = 2 + sin 2 θ tan 2 θ = 1 2 θ = π 4 , 5 π 4 , 9 π 4 , 13 π 4 θ = π 8 , 5 π 8 , 9 π 8 , 13 π 8 From the point of intersection, we can say that the common area say A 3 between the two curves will be calculated as the sum of the area of first curve for π 8 ≤ θ ≤ 5 π 8 and the second curve for 5 π 8 ≤ θ ≤ 9 π 8 . The curve r = 2 + sin 2 θ is symmetrical about the line θ = π 4 and θ = 3 π 4 . So, the area A 1 will be four times area between θ = π 4 and θ = 3 π 4 . The second curve r = 2 + cos 2 θ is symmetrical about both the axes. So, the area A 2 will be four times area n the first quadrant for 0 ≤ θ ≤ π 2 . Now, the area A 1 of the first curve r = 2 + sin 2 θ will be evaluated as : A 1 = 4 [ 1 2 ∫ π 4 3 π 4 ( 2 + sin 2 θ ) 2 d θ ] = 2 ∫ π 4 3 π 4 ( 4 + 4 sin 2 θ + sin 2 2 θ ) d θ = 2 ∫ π 4 3 π 4 ( 9 2 + 4 sin 2 θ − 1 2 cos 4 θ ) d θ = 2 [ 9 2 θ − 2 cos 2 θ − 1 8 sin 4 θ ] π 4 3 π 4 = 9 π 2 And, the area A 2 of the second curve r = 2 + cos 2 θ will be evaluated as: A 2 = 4 [ 1 2 ∫ 0 π 2 ( 2 + cos 2 θ ) 2

3rd Edition

ISBN: 9781464114885

Author: Jon Rogawski, Colin Adams

Publisher: W. H. Freeman

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Question

Chapter 11.4, Problem 22E

To determine

To calculate:

The area of the region that lies inside one but not both of the curves $r = 2 + \sin 2 θ$ and $r = 2 + \cos 2 θ$ .

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Chapter 11.4, Problem 21E

Chapter 11.4, Problem 23E

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1. Abel's Theorem. The goal in this problem is to prove Abel's theorem by following a series of steps (each step must be justified). Theorem 0.1 (Abel's Theorem). If y1 and y2 are solutions of the differential equation y" + p(t) y′ + q(t) y = 0, where p and q are continuous on an open interval, then the Wronskian is given by W (¥1, v2)(t) = c exp(− [p(t) dt), where C is a constant that does not depend on t. Moreover, either W (y1, y2)(t) = 0 for every t in I or W (y1, y2)(t) = 0 for every t in I. 1. (a) From the two equations (which follow from the hypotheses), show that y" + p(t) y₁ + q(t) y₁ = 0 and y½ + p(t) y2 + q(t) y2 = 0, 2. (b) Observe that Hence, conclude that (YY2 - Y1 y2) + P(t) (y₁ Y2 - Y1 Y2) = 0. W'(y1, y2)(t) = yY2 - Y1 y2- W' + p(t) W = 0. 3. (c) Use the result from the previous step to complete the proof of the theorem.

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Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2-t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = 1+t, y = 2 − t, z = 4 - 3t and is parallel to the plane 5x + 2y + z = 1. (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1+t, and z = 2-t. (e) The plane that contains the lines L₁: x = 1 + t, y = 1 − t, z = 2t and L2 : x = 2 − s, y = s, z = 2.

Chapter 11 Solutions

Calculus: Early Transcendentals

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 6PQ Ch. 11.1 - Prob. 7PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E

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The area of the region that lies inside one but not both of the curves r = 2 + sin 2 θ and r = 2 + cos 2 θ .

To calculate: The area of the region that lies inside one but not both of the curves r=2+sin2θ and r=2+cos2θ.

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Chapter 11 Solutions

To calculate:

The area of the region that lies inside one but not both of the curves $r = 2 + \sin 2 θ$ and $r = 2 + \cos 2 θ$ .