中山大学珠海校区2010学年度第二学期10级高等数学一期中考试题及参考答案

珠海校区2010学年度第二学期10级高等数学一期中考试题及参考答案

完成以下各题,每题10分.考试时间90分钟.

yy1. 求满足条件du?(e?sinx)dx?(xe?cosy)dy的函数u(x,y).

)e?解 P(x,y?ysxinQx,y(?,xe)?y?Py?Qyco?se,?

?y?x,故积分与路径无关,于是

u(x,y)??(e?sinx)dx??(xey?cosy)dy00x0y?xe?siny?cosx?C.2.计算累次积分: I??0dx?x211y

xy1?yy03dy.

dx解:I??dx?20x11xy1?yy03dy??dy?01xy1?y3??10x2?31?y2y11y2dy??dy3021?y2?1.3

111d(1?y3)13???1?y?306301?y3.若D?{(x,y)x?y?1},计算二重积分I???(x?y)dxdy.

D解: 积分区域关于两个坐标轴都对称,且被积函数关于x,y均为偶函数,故如记

D1??(x,y)(x,y)?D,x?0,y?0?

I???(x?y)dxdy?4??(x?y)dxdy

DD1?4?dx?011?x014(x?y)dy?4?[x(1?x)?(1?x)2]dx?.

0231C2222x?yds,其中C是圆周x?y?2x. 4. 求第一型曲线积分I???解: 用极坐标:圆周的方程为r?2cos?,故参数方程为x?2cos2?,y?2cos?sin?.

ds?(dx)2?(dy)2?(2sin2?)2?(2cos2?)2d??2d?

????22222I??x?yds?2?2cos??2d??4?????cos?d??8.

C22225.若C是上半圆周x?y?9,y?0,方向由点(3,0)到点(?3,0),求第二型曲线积分

22I?ydx?xdy. ?C解: 圆周的极坐标方程为:x?3cos?,y?3sin?,0????.故 I???0[(3sin?)2(?3sin?)?(3cos?)2(3cos?)]d?

?0?27?(cos3??sin3?)d?

?? 422???27?(1?sin?)dsin???(1?cos?)dcos??27?(?)??36.??0?0?36. 已知函数f(x)连续,求证;0f(x)dx?a?ax21?af(y)dy??f(x)dx?.

??0?2?证明;显然?a0f(x)dx?f(y)dy??f(x)dx?f(y)dy00xaaa2xaa ????????f(x)dx?f(y)dy??f(x)dx.?????0???0???0?而变换积分次序后再换积分变量字母,有?a0f(x)dx?f(y)dy??f(y)dy?f(x)dx??f(x)dx?f(y)dyx0000aayax

于是

?a0f(x)dx?ax21?a?f(y)dy??f(x)dx.证毕.

??0?2?xaxdy,?f(x)d?证法2: 记F(x)??0f(y)则0F(于是)a.

aa?a0f(x)dx?f(y)dy??f(x)[F(a)?F(x)]dx?F(a)?f(x)dx??F(x)f(x)dx

x000aa?F2(a)??a01F(x)dF(x)?F2(a)?F2(x)2a02121?a?F(a)??f(x)dx?.

??0?22?2222z?2?x?y与z?x?y所围立体的体积. 7. 求由曲面

22解: 立体在xOy坐标面的投影为区域D?{(x,y)x?y?1}.故

V????dV???dxdy?2

?D2?12002?(x2?y2)x?y2dz?2??(1?x2?y2)dxdyD?2?d??(1?r)rdr??.?

228.求三重积分I????(y?sinz)dV,其中?是由锥面z?x?y与平面z??所围的区域.

解: 由对称性,

???ydV?0.故如记区域在

?xOy面的投影区域为D,则

I????sinzdV???rdrd??sinzdz?Dr???d??(1?cosr)rdr??3?4?.002??xdy?(y?1)dx(y?1)2?29.求曲线积分I????x2?(y?1)2,其中L方程为x?2?1,逆时针方向.

L?(y?1)x?P(y?1)2?x2?QP(x,y)?,Q(x,y)?,?2?, 解: 222222x?(y?1)x?(y?1)?y[x?(y?1)]?x由于点(0,1)位于L+所围区域(记为D)内,作圆周C+: x2+y2=r2,则由格林公式,

I?xdy?(y?1)dx?0, 22x?(y?1)(L?C)???22222?rcos??rsin?xdy?(y?1)dxxdy?(y?1)dxI??d??2?.2??x2?(y?1)2????x2?(y?1)2??0rLC10.计算曲面积分I????S?ezx2?y2dxdy,其中S?为锥面z?x2?y2及平面z?1,z?2所围

立体的表面,取外侧. 解: P?Q?0,R?ezezx?y22，记S所围的区域为?,由高斯公式,

I????S?2???P?Q?R?ezdxdy???????dV ?dV????2222?x?y?z?x?yx?y???2z2?2ezrdr??d??zezdz?2?(z?1)ez01r21 ??0d??1dz?0?2e2?.

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