Rog_Sec_15_4

Rog_Sec_15_4 - (8/28/10) Math 20C. Lecture Examples Section...

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Unformatted text preview: (8/28/10) Math 20C. Lecture Examples Section 15.4. Integrals in polar coordinates† y dx dy as an iterated integral in polar coordinates, where (a) Express Example 1 R R is the region bounded by the x-axis and the half cardioid r = 1 + cos θ, 0 ≤ θ ≤ π in Figure 1. (b) Evaluate the integral. FIGURE 1 r =1+cos θ θ =π r2 sin θ dr dθ (b) y dx dy = Answer: (a) θ =π (a) θ =π r =0 r =1+cos θ θ =π r =1+cos θ r2 sin θ dr dθ (r sin θ)r dr dθ = θ =0 r =0 θ =0 θ =π 13 3r = r =0 θ =0 1 3 † Lecture 4 1 12 (1 + cos θ ) π 0 sin θ dθ • 1 (1 + cos θ)3 sin θ dθ = − 3 1 1 u3 dθ = − 12 u3 + C = − 12 (1 + cos θ)4 + C • y dx dy = R 3 1 3 (1 + cos θ ) dθ = u = 1 + cos θ, du = − sin θ dθ • 1 3 r =0 θ =π r =1+cos θ sin θ θ =0 = R (r sin θ)r dr dθ θ =0 (b) 4 3 r =1+cos θ y dx dy = R r =0 θ =0 R SOLUTION: y dx dy = 1 = − 12 (04 − 24 ) = notes to accompany Section 15.4 of Calculus by Rogawski 1 4 3 (1 + cos θ)3 (− sin θ dθ) Math 20C. Lecture Examples (8/28/10). Section 15.4, p. 2 x2 + y2 + 1 on the circle R: x2 + y2 ≤ 1. Find the average value of f (x, y) = Example 2 2 Answer: [Average value] = 3 (23/2 − 1) SOLUTION: [Area of R] = π (1)2 = π • Figure A2 Figure A2 [Average value] = 1 [Area of R] 1 = π = = = R θ =2π 1 π r =0 θ =2π 1 π r =1 dθ θ =0 2π π u = 1 + r2 , du = 2r dr • 2 2 3/2 3u r =1 r2 + 1 r dr dθ θ =0 r =1 (1 + r2 )1/2 r dr r =0 (1 + r2 )1/2 r dr r =0 r =1 r =0 u1/2 du = R x2 + y 2 + 1 dx dy =2 = x2 + y 2 + 1 dx dy (1 + r2 )1/2 r dr • (1 + r2 )1/2 r dr = (1 + r2 )1/2 (2r dr) 2 + C = 3 (1 + r2 )3/2 + C • [Average value] = = 2 3/2 2 3 (1 + r ) 2 3/2 3 (2 − 1) r =1 r =0 Section 15.4, p. 3 Math 20C. Lecture Examples (8/28/10). Evaluate Example 3 R xy dx dy, where R: 0 ≤ y ≤ xy dx dy = Answer: √ x − x2 1 24 R SOLUTION: √ y = x − x2 is the semicircle in Figure A3 with the polar equation r = cos θ, 0 ≤ θ ≤ 1 π • 2 θ =π/2 r =cos θ xy dx dy = (r cos θ)(r sin θ) r dr dθ R r =0 θ =0 θ =π/2 r =cos θ r3 cos θ sin θ dr dθ = θ =0 r =0 θ =π/2 14 4 r cos θ sin θ = θ =0 r =cos θ dθ r =0 θ =π/2 = cos5 θ sin θ dθ 1 4 θ =0 = 1 4 − 1 6 cos6 θ 1 1 = − 4 (0 − 6 ) = θ =π/2 θ =0 1 24 (using u = cos θ, du = − sin θ dθ) y √ y = x − x2 r = cos θ 1 2 Figure A3 1 x Math 20C. Lecture Examples (8/28/10). Section 15.4, p. 4 More practice Example 4 (x2 + y2 )−2 dx dy where R is defined Use polar coordinates to evaluate R by the inequalities 2 ≤ x2 + y2 ≤ 4. θ =2π (x2 + y 2 )−2 dx dy = Answer: 4 √ −4 Example 5 √ What is the value of 4 √ Answer: −4 √ r= 2 θ =0 R − 16−x2 √ 16−x2 2 e−x −y 2 r =2 16−x2 − r−3 dr dθ = e−x 2 −y2 1 π 4 dy dx? 16−x2 2 e−x dy dx = −y 2 θ =2π r =4 R 2 e−r r dr dx dy = θ =0 r =0 = π (1 − e−16 ) (Here R is the disk of radius 4 centered at the origin.) Interactive Examples Work the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡ Section 16.3: Examples 1–4 ‡ The chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sections of the textbook for the course. ...
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