关于数学分析的计算题I(积分)

不定积分与定积分

$\bf计算：$$\int {\frac{1}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$

1

$\bf计算：$$\int {\frac{{1 + \sin x}}{{1 - \cos x}}} {e^{ - x}}dx$

1

$\bf计算：$$I\left( {m,n} \right) = \int_0^1 {{x^m}{{\left( {\ln x} \right)}^n}dx} $

1

$\bf计算：$$\int_0^{2\pi } {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} $

1

$\bf计算：$$\int_0^1 {\frac{{\ln \left( {1 + x} \right)}}{{1 + {x^2}}}dx} $

1   2   3

$\bf计算：$

反常积分与含参积分

$\bf计算：$$I = \int_0^{ + \infty } {{e^{ - px}}\frac{{\sin bx - \sin ax}}{x}dx} \left( {p > 0,b > a} \right)$

1

$\bf计算：$$\int_0^{ + \infty } {\frac{1}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} dx$

1

$\bf计算：$$\int_0^{ + \infty } {\frac{{1 - {e^{ - t}}}}{t}\sin tdt} $

1

$\bf计算：$

重积分

$\bf计算：$设$D = \left\{ {\left( {x,y} \right)| - 1 \le x \le 1,0 \le y \le 1} \right\}$，计算$\iint\limits_D {\sqrt {\left| {y - {x^2}} \right|} }dxdy$

1

$\bf计算：$设$D$由$y = x,y = 0,x = \frac{\pi }{2}$围成，计算$\iint\limits_D {\left| {\cos \left( {x + y} \right)} \right|dxdy}$

1

$\bf计算：$设$\Omega $由锥面${x^2} + {y^2} = {z^2}$和$z=2$所围成，计算$\iiint\limits_\Omega  {\sqrt {{x^2} + {y^2}} dxdydz}$

1

$\bf计算：$设$V$为单位球${x^2} + {y^2} + {z^2} \leqslant 1$，$a,b,c$为不全为零的常数，计算$I = \iiint\limits_V {\cos \left( {ax + by + cz} \right)dxdydz}$

$\bf计算：$

曲线积分

$\bf计算：$设$c$为${x^2} + {y^2} + {z^2} = 9$与$x + y + z = 0$的交线，计算$\int_c {xyds} $

1

$\bf计算：$设$L$为${x^2} + {y^2} = 1$，取逆时针方向，计算$\oint_L {\frac{{ - ydx + xdy}}{{4{x^2} + {y^2}}}} $

1

$\bf计算：$

曲面积分

$\bf计算：$设锥面$S$为${x^2} + {y^2} = {\left( {1 - z} \right)^2},0 \leqslant z \leqslant 1$，计算$\iint\limits_S {\frac{{{x^3} + {y^3} + {z^3}}}{{1 - z}}dS}$

1

$\bf计算：$设$S$为柱面${x^2} + {z^2} = 2az\left( {a > 0} \right)$被锥面$z = \sqrt {{x^2} + {y^2}} $所截取的有限部分，计算$\iint\limits_S {\left( {x + z} \right)dS}$

1

$\bf计算：$设$S$为球面${x^2} + {y^2} + {\left( {z - a} \right)^2} = {a^2}$中满足${x^2} + {y^2} \leqslant ay$与$z \leqslant a$那部分的下侧$(a>0)$，计算$\iint\limits_S {{x^2}dydz + zdxdy}$

1

$\bf计算：$设$S$为抛物面${x^2} + {y^2} = z$位于$z=0,z=1$之间的部分，取外侧，计算$\iint\limits_S {2xydydz - {y^2}dzdx - {x^2}dxdy}$

1

$\bf计算：$设$f(x,y,z)$表示从原点到椭球面$\Sigma :\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$上$p(x,y,z)$处的切平面的距离，计算$\iint\limits_\Sigma  {\frac{{dS}}{{f\left( {x,y,z} \right)}}}$

1

$\bf计算：$设$f\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}{1 - {x^2} - {y^2} - {z^2},{x^2} + {y^2} + {z^2} \leqslant 1} \\ {0{\text{ ,}}{x^2} + {y^2} + {z^2} > 1} \end{array}} \right.$，计算$F\left( t \right) = \iint\limits_{x + y + z = t} {f\left( {x,y,z} \right)dS}$

1

$\bf计算：$

附录

$\bf命题：$设$P(x,y),Q(x,y)$在$R^2$上有连续的偏导数，且对任何一个圆周$C$，有$\int_C {P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0} $，证明：$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}$

1

$\bf命题：$是$S$为球面${x^2} + {y^2} + {z^2} = 1,f$为连续函数，$a,b,c$为常数，证明$\bf{Poisson}公式$：\[\iint\limits_S {f\left( {ax + by + cz} \right)dS} = 2\pi \int_{ - 1}^1 {f\left( {\sqrt {{a^2} + {b^2} + {c^2}} u} \right)du} \]

1

$\bf命题：$

关于数学分析的计算题I(积分)

原文：http://www.cnblogs.com/ly142857/p/3747080.html