高数打卡04

1. 线 r = a ( 1 + c o s θ ) ( a > 0 , 0 θ 2 π ) . 1.求心脏线r=a(1+cos\theta)(a>0,0\leqslant\theta \leqslant2\pi)所围区域之面积.
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S = 0 2 π d θ 0 a ( 1 + cos θ ) r d r = 1 2 0 2 π a 2 ( 1 + cos θ ) 2 d θ = ( 1 2 a 2 θ + a 2 sin θ + 1 4 a 2 θ + 1 8 a 2 sin 2 θ ) 0 2 π = 3 2 π a 2 \begin{aligned} &S=\int_{0}^{2 \pi} d \theta \int_{0}^{a(1+\cos \theta)} r d r \\ &=\frac{1}{2} \int_{0}^{2 \pi} a^{2}(1+\cos \theta)^{2} d \theta \\ &=(\frac{1}{2} a^{2} \theta+a^{2} \sin \theta+\frac{1}{4} a^{2} \theta+\frac{1}{8} a^{2} \sin 2\theta)|_{0} ^{2 \pi} \\ &=\frac{3}{2} \pi a^{2} \end{aligned}
2. 线 r 2 = 4 c o s 2 θ . 2.求双纽线r^2=4cos2\theta所围区域的面积.
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S = 4 0 π 4 d θ 0 2 cos 2 θ r d r = 4 0 π 4 2 cos 2 θ d θ = 4 \begin{aligned} &S=4 \int_{0}^{\frac{\pi}{4}} d \theta \int_{0}^{2 \sqrt{\cos 2 \theta}} r d r\\ &=4 \int_{0}^{\frac{\pi}{4}} 2 \cos 2 \theta d \theta\\ &=4 \end{aligned}
3. 线 r = 1 + s i n θ ( 0 θ 2 π ) . 3.求心脏线r=1+sin\theta(0\leqslant\theta \leqslant2\pi)所围区域中位于第一象限部分的面积.
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S = 0 π 2 d θ 0 1 + sin θ r d r = 1 2 0 π 2 ( 1 + sin θ + sin 2 θ ) d θ = 1 + 3 8 π \begin{aligned} &S=\int_{0}^{\frac{\pi}{2}} d \theta \int_{0}^{1+\sin \theta} r d r\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(1+\sin \theta+\sin ^{2} \theta\right) d \theta\\ &=1+\frac{3}{8} \pi \end{aligned}

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转载自blog.csdn.net/qq_45645641/article/details/105574986