双曲抛物面$z=xy$被圆柱面${{x}^{2}}+{{y}^{2}}={{a}^{2}}$截下部分的面积为( )
A: $\frac{2\pi }{3}[{{(1+a)}^{3/2}}+1]$
B: $\frac{2\pi }{3}[{{(1+a)}^{3/2}}-1]$
C: $\frac{2\pi }{3}[{{(1+a)}^{2/3}}+1]$
D: $\frac{2\pi }{3}[{{(1+a)}^{2/3}}-1]$
A: $\frac{2\pi }{3}[{{(1+a)}^{3/2}}+1]$
B: $\frac{2\pi }{3}[{{(1+a)}^{3/2}}-1]$
C: $\frac{2\pi }{3}[{{(1+a)}^{2/3}}+1]$
D: $\frac{2\pi }{3}[{{(1+a)}^{2/3}}-1]$
举一反三
- 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: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- 设`\n`阶方阵`\A`满足`\|A| = 2`，则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于（ ） A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]
- 微分方程$y' = \sqrt{x},y(1)=0$的解为 A: $ \frac{2}{3} x^{\frac{3}{2}} + C $ B: $ \frac{2}{3} x^{\frac{3}{2}} -\frac{2}{3} $ C: $ x^{\frac{3}{2}}-1 $ D: $ x^{\frac{3}{2}}+C $
- 这时线圈平面法线方向与该处磁感强度的方向的夹<br/>角为____________________． A: `\frac{1}{3}\pi` B: `\frac{1}{6}\pi` C: `\frac{1}{2}\pi` D: `\frac{2}{3}\pi`