# 104 - 6 sin θ θ 1 2 sin 2 θ ²³ ³ ³ ³ 5 π 4 π 4 ±...

This preview shows pages 1–3. Sign up to view the full content.

MATH 153 Selected Solutions for § 10.4 Exercise 8 : Find the area of the shaded region. - 0.5 0.0 0.5 - 0.5 0.0 0.5 Figure 1. r = sin 2 θ . Solution : We ﬁrst ﬁnd the θ values where the curve hits the origin by solving for r = 0. But r = 0 if sin 2 θ = 0, or if 2 θ = 0 . Thus, the values we’re going to use to integrate are 0 and π/ 2. A = Z π/ 2 0 1 2 r 2 = 1 2 Z π/ 2 0 sin 2 2 θ dθ = 1 2 Z π/ 2 0 1 - cos 4 θ 2 = θ 4 - sin 4 θ 16 ± ± ± ± π/ 2 0 = π/ 2 4 - sin 2 π 16 - ( 0 4 - sin 0 16 ) = π 8 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 Exercise 32 : Find the area of the region that lies inside both curves. r = 3 + 2 cos θ, r = 3 + 2 sin θ - 4 - 2 0 2 4 - 4 - 2 0 2 4 Figure 2. The two curves with desired area shaded. Solution : Setting the two curves equal to each other, we see that they in- tersect when sin θ = cos θ , or at θ = π/ 4 , 5 π/ 4. Thus, at these θ values, we switch the function that governs the boundary of the area, and we thus break up the integral there. Notice we integrate from π/ 4 to 9 π/ 4 for simplicity. A = Z 5 π/ 4 π/ 4 1 2 (3 + 2 cos θ ) 2 + Z 9 π/ 4 5 π/ 4 1 2 (3 + 2 sin θ ) 2 = Z 5 π/ 4 π/ 4 9 2 + 6 cos θ + 2 cos 2 θ dθ + Z 9 π/ 4 5 π/ 4 9 2 + 6 sin θ + 2 sin 2 θ dθ = Z 5 π/ 4 π/ 4 9 2 + 6 cos θ + 1 + cos 2 θ dθ + Z 9 π/ 4 5 π/ 4 9 2 + 6 sin θ + 1 - cos 2 θ dθ = ± 9 θ 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + 6 sin θ + θ + 1 2 sin 2 θ ²³ ³ ³ ³ 5 π/ 4 π/ 4 + ± 9 θ 2-6 cos θ + θ-1 2 sin 2 θ ²³ ³ ³ ³ 9 π/ 4 5 π/ 4 = ± 45 π 8-3 √ 2 + 5 π 4 + 1 2 ²-± 9 π 8 + 3 √ 2 + π 4 + 1 2 ² + ± 81 π 8-3 √ 2 + 9 π 4-1 2 ²-± 45 π 8 + 3 √ 2 + 5 π 4-1 2 ² = 11 π-12 √ 2 3 Exercise 46 : Find the exact length of the polar curve. r = e 2 θ , ≤ θ ≤ 2 π 20000 40000 60000 80000 100000 120000-50000-40000-30000-20000-10000 Figure 3. The curve r = θ with 0 ≤ θ ≤ 2 π . Solution : Note that in the picture above, the function grows so quickly that Mathematica refused to draw it all the way to 2 π . This is going to be a large number. L = Z 2 π s r 2 + ± dr dθ ² 2 dθ = Z 2 π p e 4 θ + 4 e 4 θ dθ = Z 2 π √ 5 e 2 θ dθ = √ 5 2 e 2 θ ³ ³ ³ ³ ³ 2 π = √ 5 2 e 4 π-√ 5 2 ’ 320 , 597 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

104 - 6 sin θ θ 1 2 sin 2 θ ²³ ³ ³ ³ 5 π 4 π 4 ±...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online