计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。
A: \(2\pi {a^7}\)
B: \(2\pi {a^6}\)
C: \(2\pi {a^5}\)
D: \(2\pi {a^8}\)
A: \(2\pi {a^7}\)
B: \(2\pi {a^6}\)
C: \(2\pi {a^5}\)
D: \(2\pi {a^8}\)
举一反三
- 计算\({\oint_L {({x^2} + {y^2})} ^n}ds\),其中\(L\)为圆周\(x = a\cos t\),\(y=asint\)\((0 \le t \le 2\pi )\)。 A: \(2\pi {a^{n + 1}}\) B: \(2\pi {a^{2n + 1}}\) C: \(\pi {a^{n + 1}}\) D: \(2\pi {a^{n + 1}}\)
- 已知\(L\)为圆周 \(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\),则\({\oint_L {({x^2} + {y^2})} ^n}ds{\rm{ = }}\) ( ). A: \(2\pi {a^{2n + 1}}\) B: \(2\pi {a^{2n - 1}}\) C: \(\pi {a^{2n + 1}}\) D: \(\pi {a^{2n - 1}}\)
- 曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 旋轮线$x=a(t-\sin t),y=a(1-\cos t)$的一拱($0 \le t \le 2 \pi$)的绕$x$轴旋转得到的立体的体积为 A: $\pi a^3$ B: $\frac{32}{105} \pi a^3$ C: $\pi a^2$ D: $\frac{32}{105} \pi a^2$
- 应用格林公式可计算星形线$x=a\cos^3t$, $y=a\sin^3 t$所围的平面面积为 A: $\pi a^2$ B: $\frac{3}{4}\pi a^2$ C: $\frac{3}{8}\pi a^2$ D: $\frac{3}{16}\pi a^2$