二重積分的極座標形式

Section 3 二重積分的極座標形式

3.1 (1)x=4cos^π₃,y=4sin^π₃ (2)x=−1cos^5π4 ,y=−1sin^5π4

(3)x=0cos2π,y=0sin2π (4)r=1²+ 0²,tanθ=⁰₁ (5)r=0²+ 1²,tanθ=¹₀

(6)r=(−1)²+ (−1)²,tanθ=⁻¹

−1

3.2 圖在下一頁 (1)r=4cosθ

=> r²=4rcosθ

=> x²+ y²=4x

=> (x − 2)²+ y²=2

(2)r=2sinθ

=> r²=2rsinθ

=> x²+ y²=2y

=> x²+ (y − 1)²=1

(2)rcos(θ − ^π4)=1

=> r(cos^π₄cosθ+ sin^π₄sinθ)=1

=> ^√x₂ +^√y₂=1

=> x + y=√ 2

3.3 (1)x²− y²=(rcosθ)²− (rsinθ)²=r²(cos²θ− sin²θ)=r²cos2θ (2)xy=(rcosθ)(rsinθ)=r²sinθcosθ=¹₂r²sin2θ

(3)x+y-1=rcosθ + rsinθ − 1 (4)e^−(x2+y2)=e^−r2

Figure 1: 3.2(1)

Figure 2: 3.2(2)

Figure 3: 3.2(3)

3.4 (1)^R₀^πR₂⁴rsinθrdrdθ

=^R₀^π(¹₃r³sinθ)|⁴2dθ

=^R₀^{π 56}₃cosθdθ

=⁵⁶₃ (−cosθ|^π0)

=¹¹²₃

(2)^R₀^πR₀³(√

1 + 2r²+ θ)rdrdθ

=^R₀^πR₀³r√

1 + 2r²+ rθdrdθ

=^R₀^π(¹₆(1 + 2r²)³² +¹₂r²θ)|³0dθ

=^R₀^π(¹₆19³² +⁹₂θ− ¹₆)dθ

=¹₆(19³² − 1)θ +⁹₄θ²

=¹₆π(19³² − 1) + ⁹₄π² (3)^R

π 4

0

R_cosθ

sinθ (√

1 − r²tanθ)rdrdθ

=^R

π 4

0 (−¹₃(1 − r²)³²tanθ)^cosθ_sinθdθ

=^R

π 4

0 (sin³θ− cos³θ)(tanθ)dθ

=^R

π 4

0 (sinθ − cosθ)(tanθ + sin²θ)dθ

=^R

π 4

0 sinθtanθ+ sin³θ− sinθ − sin²θcosθdθ

=^R

π 4

0 secθ− cosθ + sinθ(−cos²θ) − sin²θcosθdθ

=(ln|secθ + tanθ| − sinθ + ¹₃cos³θ− ¹₃sin³θ)

π 4

0

=ln(1 +√

2) −^√1₂ − ¹₃ (4)^R

π 2

0

R^√sin2θ 0 rdrdθ

=^R

π 2

0 (¹₂r²)|^√0^sin2θdθ

=^R

π 2

0 1

2sin2θdθ

=−¹4cos2θ|

π 2

0

=¹₂

3.5 (1)^R

π 2

0

R_sin2θ

0 rdrdθ

=^R

π 2

0 1

2r²)^sin2θ₀ dθ

=^R

π 2

0 1

2(sin2θ)²dθ

=^R

π 2

0 1

4(1 − cos4θ)dθ

=(¹₄θ− 16¹sin4θ)

π 2

0

(2)^R

3π 2 π 2

R_1−cosθ

1 rdrdθ

=^R

3π 2 π

2 (¹₂r²)^1−cosθ₁ dθ

=^R

3π 2 π 2

1

2cos²θ− cosθdθ

=^R

3π 2 π 2

1

4(1 + cos2θ) − cosθdθ

=(¹₄(θ + ¹₂sin2θ) − sinθ)

3π 2 π 2

=^π₄ + 2

(3)^R₀^πR₀¹e^r2rdrdθ

=^R₀^π(¹₂e^r2)|¹0dθ

=^R₀^{π 1}₂(e − 1)dθ

=¹₂(e − 1)θ|^π0

=¹₂π(e − 1)

(4)^R₀^πR₀¹ln(1 + r²)rdrdθ 設_{u=ln(1 + r}²_),dv=rdr 則_du= ^2r

1+r²dr,v=¹₂r²

=^R₀^π(¹₂r²ln(1 + r²)|¹0−^R0¹ 1

2r²× 1+r^2r2dr)dθ

=^R₀^π(¹₂ln2 −^R0¹ r³

1+r²dr)dθ

=^R₀^π(¹₂ln2 −^R0¹r− _1+r^r2dr)dθ

=^R₀^{π 1}₂ln2 − (¹₂r²− ¹₂ln(1 + r²))|¹0dθ

=^R₀^{π 1}₂ln2 − (¹₂ − ¹₂ln2)dθ

=^R₀^π(ln2 −¹₂)dθ

=(ln2 − ¹₂)θ|^π0

=π(ln2 − ¹₂) (5)^R₀^πR

1 2

0 r²sinθdrdθ

=^R₀^π(¹₃r³sinθ)

1 2

0dθ

=^R₀^π ₂₄¹sinθdθ

=₂₄¹ (−cosθ)^π0

=₁₂¹

3.6 (1)0 ≤ x ≤ 2 ⇒ 0 ≤ r ≤ cosθ²

0 ≤ y ≤ x ⇒ 0 ≤ rsinθ ≤ rcosθ ⇒ 0 ≤ tanθ ≤ 1 ⇒ 0 ≤ θ ≤ ^π4

(2)^R

π 4

0

Rcosθ²

0 r²sinθdrdθ

=^R

π 4

0 (¹₃r³sinθ)

2 cosθ

0 dθ

=^R

π 4

0 8

3tanθsec²θdθ

=⁴₃tan²θ|

π 4

0

=⁴₃

3.8 考慮內接圓的範圍 ₍畫圖可知半徑為一):0 ≤ r ≤ 1,0 ≤ θ ≤ 2π 積分

R_2π

0

R₁

0 1

1+r²rdrdθ

=^R₀^2π(¹₂ln(1 + r²))¹₀dθ

=^R₀^{2π 1}₂ln2dθ

=(¹₂ln2)θ|^2π0

=πln2

接著_, 換考慮外接圓的範圍 ₍畫圖可知半徑為^√_{2):0 ≤ r ≤}^√2,0 ≤ θ ≤ 2π 積分

R_2π

0

R^√₂

0 1

1+r²rdrdθ

=^R₀^2π(¹₂ln(1 + r²))^√₀²dθ

=^R₀^{2π 1}₂ln3dθ

=(¹₂ln3)θ|^2π0

=πln3

由於內接圓_{⊂ Ω ⊂}外接圓_, 所以由習題_1.6可知 πln2 <^{R R}_{Ω 1+x}¹²_+y²dA < πln3