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 xaxis 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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 Spring '08
 Helton
 Calculus, Integrals, Cartesian Coordinate System, Polar Coordinates, Coordinate system, dr dθ, Polar coordinate system, dx dy

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