题目
【题目】计算由摆线 x=a(t-sint) y=a(1-cost) 相应于0≤t≤2π 的一拱与直线y=0所围成的图形分别绕x轴、y轴旋转而成的旋转体的体积.
【题目】计算由摆线 x=a(t-sint) y=a(1-cost) 相应于0≤t≤2π 的一拱与直线y=0所围成的图形分别绕x轴、y轴旋转而成的旋转体的体积.
题目解答
答案
【解析】按旋转体的体积公式,所述图形绕轴旋转而成的旋转体的体积为V_x=∫_0^(2πa)πy^2(x)dx=π∫_0^(2π)a^2(1-cost)^2⋅a(1-cost)dt (1-3cost+3cos2t -cos3t) dt = 5π2a3.所述图形绕y轴旋转而成的旋转体的体积可看成平面图形OABC与OBC(图6-12)分别绕y轴旋转而成的旋转体的体积之差因此所求的体积为V_v=∫_0^(2a)πx_2^2(y)dy-∫_0^(2a)(πx_1^2)(y)dy =π∫_(2π)^π(a^2(t-sint)^2⋅asintdt-π∫_0^x)(a^2(t-sint)^2⋅asintdt) =-πa^3∫_0^(2π)(t-sint)^2sintdt=6π^3a^3y2aCBx=x1(y)x=x2(y)A0πa2πa x图6-12
解析
步骤 1：计算绕x轴旋转的体积
根据旋转体的体积公式，绕x轴旋转的体积为：
\[ V_x = \int_{0}^{2\pi} \pi y^2(x) dx \]
将摆线的参数方程代入，得到：
\[ V_x = \pi \int_{0}^{2\pi} a^2(1-\cos t)^2 \cdot a(1-\cos t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-\cos t)^3 dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + 3\cos^2 t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + \frac{3}{2}(1+\cos 2t) - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + \frac{3}{2} + \frac{3}{2}\cos 2t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (\frac{5}{2} - 3\cos t + \frac{3}{2}\cos 2t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \left[ \frac{5}{2}t - 3\sin t + \frac{3}{4}\sin 2t - \frac{1}{4}\sin 4t \right]_{0}^{2\pi} \]
\[ V_x = \pi a^3 \left[ \frac{5}{2} \cdot 2\pi \right] = 5\pi^2 a^3 \]
步骤 2：计算绕y轴旋转的体积
绕y轴旋转的体积可看成平面图形OABC与OBC分别绕y轴旋转而成的旋转体的体积之差。因此所求的体积为：
\[ V_y = \int_{0}^{2a} \pi x_2^2(y) dy - \int_{0}^{2a} \pi x_1^2(y) dy \]
\[ V_y = \pi \int_{2\pi}^{\pi} a^2(t-\sin t)^2 \cdot a\sin t dt - \pi \int_{0}^{2\pi} a^2(t-\sin t)^2 \cdot a\sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t-\sin t)^2 \sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t^2 - 2t\sin t + \sin^2 t) \sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t^2\sin t - 2t\sin^2 t + \sin^3 t) dt \]
\[ V_y = -\pi a^3 \left[ -\frac{1}{3}t^3\cos t + \frac{2}{3}t^3 - \frac{1}{3}t^3\cos t + \frac{1}{3}t^3 - \frac{1}{4}\cos^4 t \right]_{0}^{2\pi} \]
\[ V_y = -\pi a^3 \left[ -\frac{1}{3}(2\pi)^3\cos 2\pi + \frac{2}{3}(2\pi)^3 - \frac{1}{3}(2\pi)^3\cos 2\pi + \frac{1}{3}(2\pi)^3 - \frac{1}{4}\cos^4 2\pi \right] \]
\[ V_y = 6\pi^3 a^3 \]
根据旋转体的体积公式，绕x轴旋转的体积为：
\[ V_x = \int_{0}^{2\pi} \pi y^2(x) dx \]
将摆线的参数方程代入，得到：
\[ V_x = \pi \int_{0}^{2\pi} a^2(1-\cos t)^2 \cdot a(1-\cos t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-\cos t)^3 dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + 3\cos^2 t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + \frac{3}{2}(1+\cos 2t) - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (1-3\cos t + \frac{3}{2} + \frac{3}{2}\cos 2t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \int_{0}^{2\pi} (\frac{5}{2} - 3\cos t + \frac{3}{2}\cos 2t - \cos^3 t) dt \]
\[ V_x = \pi a^3 \left[ \frac{5}{2}t - 3\sin t + \frac{3}{4}\sin 2t - \frac{1}{4}\sin 4t \right]_{0}^{2\pi} \]
\[ V_x = \pi a^3 \left[ \frac{5}{2} \cdot 2\pi \right] = 5\pi^2 a^3 \]
步骤 2：计算绕y轴旋转的体积
绕y轴旋转的体积可看成平面图形OABC与OBC分别绕y轴旋转而成的旋转体的体积之差。因此所求的体积为：
\[ V_y = \int_{0}^{2a} \pi x_2^2(y) dy - \int_{0}^{2a} \pi x_1^2(y) dy \]
\[ V_y = \pi \int_{2\pi}^{\pi} a^2(t-\sin t)^2 \cdot a\sin t dt - \pi \int_{0}^{2\pi} a^2(t-\sin t)^2 \cdot a\sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t-\sin t)^2 \sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t^2 - 2t\sin t + \sin^2 t) \sin t dt \]
\[ V_y = -\pi a^3 \int_{0}^{2\pi} (t^2\sin t - 2t\sin^2 t + \sin^3 t) dt \]
\[ V_y = -\pi a^3 \left[ -\frac{1}{3}t^3\cos t + \frac{2}{3}t^3 - \frac{1}{3}t^3\cos t + \frac{1}{3}t^3 - \frac{1}{4}\cos^4 t \right]_{0}^{2\pi} \]
\[ V_y = -\pi a^3 \left[ -\frac{1}{3}(2\pi)^3\cos 2\pi + \frac{2}{3}(2\pi)^3 - \frac{1}{3}(2\pi)^3\cos 2\pi + \frac{1}{3}(2\pi)^3 - \frac{1}{4}\cos^4 2\pi \right] \]
\[ V_y = 6\pi^3 a^3 \]