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: \(\pi {a^{2n - 1}}\)
A: \(2\pi {a^{2n + 1}}\)
B: \(2\pi {a^{2n - 1}}\)
C: \(\pi {a^{2n + 1}}\)
D: \(\pi {a^{2n - 1}}\)
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: \(2\pi {a^{n + 1}}\)
- 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: \(2\pi {a^8}\)
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 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: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 中国大学MOOC: 下列程序的运行结果是（ ）。x=0:pi/100:2*pi;for n=1:2:10 plot(n*sin(x),n*cos(x)) hold onendaxis square