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极坐标面积计算

Créé le : 27 janvier 2025

Répondu en utilisant O1 par Chat01

Question

求 r^2 = 2a^2cos(2θ) 围成区域的面积

Réflexion

Piecing together the area

I'm examining the polar equation r2=2a2cos⁡(2θ)r^2 = 2a^2 \cos(2θ) to find the angular range where cos⁡(2θ)≥0\cos(2θ) \geq 0. This involves analyzing the integrals within the specified range to calculate the area.

Clarifying the intervals

I’m digging into the specific intervals for θ\theta where cos⁡(2θ)≥0\cos(2θ) \geq 0, ensuring the analysis covers the principal domain of 00 to 2π2π.

Identifying valid intervals

I’m systematically testing intervals of θ\theta to determine where cos⁡(2θ)≥0\cos(2θ) \geq 0 within [0,2π][0, 2π]. This helps pinpoint the valid segments for the area calculation.

Refining the interval

I'm verifying that cos⁡(2θ)≥0\cos(2θ) \geq 0 in the interval [7π/4,2π][7π/4, 2π]. Adjusting my approach to ensure the calculations are accurate within this range.

Pinpointing valid intervals

I’ve identified the valid intervals for r2≥0r^2 \geq 0 in θ\theta as [0,π/4][0, \pi/4] and [3π/4,5π/4][3\pi/4, 5\pi/4]. Then, I’m calculating the area using these intervals.

Breaking down the integration

I’m integrating cos⁡(2θ)\cos(2θ) over [0,π/4][0, \pi/4] and [3π/4,5π/4][3\pi/4, 5\pi/4]. The antiderivative is (1/2)sin⁡(2θ)(1/2) \sin(2θ), and I evaluate these integrals to find the area.

Combining integrals

I’m integrating over the intervals [0,π/4][0, \pi/4] and [3π/4,5π/4][3\pi/4, 5\pi/4], and then factoring in (1/2)(1/2) for the area calculation.

Finalizing the area

I’m re-evaluating the integral of cos⁡(2θ)\cos(2θ) over the specified intervals, ensuring the factor (1/2)(1/2) is correctly applied, leading to the final area expression.

Mapping out the area

I’ve determined the area using the polar equation (1/2)∫r2dθ(1/2) \int r^2 dθ within the specified intervals, leading to the final expression 3a2/23a^2/2.

Verifying the result

I’ve confirmed that the final area enclosed by the curve r2=2a2cos⁡(2θ)r^2= 2 a^2 \cos(2θ) is 3a22\frac{3a^2}{2}, noting the curve's resemblance to a scaled lemniscate.

Thought about r^2 = 2a^2cos(2θ) 形状，持续 1m 56s

Réponse

下面给出详细的推导步骤。我们要计算由极坐标方程

r2=2a2cos⁡(2θ)r^2 = 2 a^2 \cos(2\theta)

所围成的区域的面积。

1. 确定 r2≥0r^2\ge 0 对应的 θ\theta 范围

因为半径平方 r2r^2 必须非负，所以需要满足

2a2cos⁡(2θ)≥0⟹cos⁡(2θ)≥0.2 a^2 \cos(2\theta) \ge 0 \quad\Longrightarrow\quad \cos(2\theta) \ge 0.

在 0≤θ<2π0 \le \theta < 2\pi 内，cos⁡(2θ)\cos(2\theta) 为非负的区间可通过“找零点+符号检测”来确定。

  • cos⁡(2θ)=0\cos(2\theta)=0 时，2θ=π2+kπ2\theta = \tfrac{\pi}{2} + k\pi。
    因而 θ=π4+kπ2.\theta = \frac{\pi}{4} + \frac{k\pi}{2}.
  • 在 0≤θ<2π0 \le \theta < 2\pi 中，零点依次是 π4,  3π4,  5π4,  7π4.\frac{\pi}{4},\;\frac{3\pi}{4},\;\frac{5\pi}{4},\;\frac{7\pi}{4}.
  • 检验相邻零点间 cos⁡(2θ)\cos(2\theta) 的正负性，可得： cos⁡(2θ)≥0在[0, π4]  ∪  [3π4, 5π4].\cos(2\theta) \ge 0 \quad \text{在} \quad [0,\, \tfrac{\pi}{4}] \;\cup\; [\tfrac{3\pi}{4},\, \tfrac{5\pi}{4}].

因此，真正有面积贡献的 θ\theta 范围是

θ∈[0, π4]  ∪  [3π4, 5π4].\theta \in [0,\, \tfrac{\pi}{4}] \;\cup\; [\tfrac{3\pi}{4},\, \tfrac{5\pi}{4}].

2. 用极坐标面积公式计算

极坐标下计算面积的公式为

Area=12∫r2(θ) dθ.\text{Area} = \frac{1}{2} \int r^2(\theta)\,d\theta.

在本题中，r2(θ)=2a2cos⁡(2θ)r^2(\theta) = 2 a^2 \cos(2\theta)。
故对满足 cos⁡(2θ)≥0\cos(2\theta)\ge0 的区间做分段积分：

Area=12[∫0π/42a2cos⁡(2θ) dθ  +  ∫π/43π/4 ⁣(负值区不计)  +  ∫3π/45π/42a2cos⁡(2θ) dθ].\text{Area} = \frac{1}{2} \biggl[ \int_{0}^{\pi/4} 2 a^2 \cos(2\theta)\,d\theta \;+\; \int_{\pi/4}^{3\pi/4}\! (\text{负值区不计}) \;+\; \int_{3\pi/4}^{5\pi/4} 2 a^2 \cos(2\theta)\,d\theta \biggr].

实际只需在区间 [0,π4][0,\tfrac{\pi}{4}] 和 [3π4,5π4][\tfrac{3\pi}{4}, \tfrac{5\pi}{4}] 积分，其他区间因为 cos⁡(2θ)<0\cos(2\theta)<0 对面积无贡献。

将常数 2a22 a^2 和系数 12\tfrac12 合并后，可以写成

Area=(12)(2a2)[∫0π/4cos⁡(2θ) dθ  +  ∫3π/45π/4cos⁡(2θ) dθ]=a2[∫0π/4cos⁡(2θ) dθ  +  ∫3π/45π/4cos⁡(2θ) dθ].\text{Area} = \left(\frac{1}{2}\right)\bigl(2 a^2\bigr) \left[ \int_{0}^{\pi/4} \cos(2\theta)\,d\theta \;+\; \int_{3\pi/4}^{5\pi/4} \cos(2\theta)\,d\theta \right] = a^2 \left[ \int_{0}^{\pi/4} \cos(2\theta)\,d\theta \;+\; \int_{3\pi/4}^{5\pi/4} \cos(2\theta)\,d\theta \right].

2.1 第一个区间 [0, π/4][0,\, \pi/4]

∫0π/4cos⁡(2θ) dθ=[12sin⁡(2θ)]0π/4=12[sin⁡(π2)−sin⁡(0)]=12×(1−0)=12.\int_{0}^{\pi/4} \cos(2\theta)\,d\theta = \left[ \tfrac{1}{2} \sin(2\theta) \right]_{0}^{\pi/4} = \frac{1}{2}\bigl[\sin(\tfrac{\pi}{2}) - \sin(0)\bigr] = \frac{1}{2}\times (1 - 0) = \frac{1}{2}.

2.2 第二个区间 [3π4, 5π4][\tfrac{3\pi}{4},\, \tfrac{5\pi}{4}]

∫3π/45π/4cos⁡(2θ) dθ=[12sin⁡(2θ)]3π/45π/4.\int_{3\pi/4}^{5\pi/4} \cos(2\theta)\,d\theta = \left[\tfrac{1}{2} \sin(2\theta) \right]_{3\pi/4}^{5\pi/4}.
  • 当 θ=3π4\theta=\tfrac{3\pi}{4} 时，2θ=3π22\theta = \tfrac{3\pi}{2}，sin⁡(3π2)=−1\sin(\tfrac{3\pi}{2})=-1。
  • 当 θ=5π4\theta=\tfrac{5\pi}{4} 时，2θ=5π22\theta = \tfrac{5\pi}{2}，sin⁡(5π2)=sin⁡(2π+π2)=1\sin(\tfrac{5\pi}{2})=\sin(2\pi+\tfrac{\pi}{2})=1。

因此

∫3π/45π/4cos⁡(2θ) dθ=12[ 1−(−1)]=12×2=1.\int_{3\pi/4}^{5\pi/4} \cos(2\theta)\,d\theta = \tfrac{1}{2}\bigl[\,1 - (-1)\bigr] = \frac{1}{2}\times 2 = 1.

2.3 总积分

将两部分结果相加：

∫0π/4cos⁡(2θ) dθ  +  ∫3π/45π/4cos⁡(2θ) dθ=12+1=32.\int_{0}^{\pi/4} \cos(2\theta)\,d\theta \;+\; \int_{3\pi/4}^{5\pi/4} \cos(2\theta)\,d\theta = \frac{1}{2} + 1 = \frac{3}{2}.

最后乘上 a2a^2：

Area=a2×32=32 a2.\text{Area} = a^2 \times \frac{3}{2} = \frac{3}{2}\,a^2.

3. 最终答案

由 r2=2a2cos⁡(2θ) 围成的区域面积为 3a22.\boxed{ \text{由 }r^2 = 2a^2\cos(2\theta)\text{ 围成的区域面积为 } \frac{3 a^2}{2}. }

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