请问积分路径怎么确定的呢?
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/15 14:20:12
![](http://img.wesiedu.com/upload/4/28/4283e91cc26f491959b2ee865f95cd49.jpg)
请问积分路径怎么确定的呢?
![请问积分路径怎么确定的呢?](/uploads/image/z/20295899-35-9.jpg?t=%26nbsp%3B%E8%AF%B7%E9%97%AE%E7%A7%AF%E5%88%86%E8%B7%AF%E5%BE%84%E6%80%8E%E4%B9%88%E7%A1%AE%E5%AE%9A%E7%9A%84%E5%91%A2%EF%BC%9F)
![](http://img.wesiedu.com/upload/5/1c/51c41160730a842c919a404e2c64f5f4.jpg)
1、首先画出二重积分区域图,两条直线,y=x,y=2x,补充二条直线x=2,y=2,求出交点坐标A(2,2),B(1,2),C(2,4),
原区域为△OAC,改变积分顺序,x=y/2,x=y,
区域由△OAB和△BAC组成,
原式=∫[0,2]dy∫[y/2,y]f(x,y)dx+∫[2,4]dy∫[2x,4]f(x,y)
2、 很明显xyz+C的全微分是yzdx+xzdy+xydz,
设P(x,y,z)=yz,Q(x,y,z)=xz,R(x,y,z)=xy,
∂P/∂x=∂Q/∂y=∂R/∂z=0,
∂P/∂x.+ ∂Q/∂y+∂R/∂z=0,
故原式是某函数u(x,y,z)的全微分,积分路径O(0,0,0)→A(x,0,0)B(x,y,0)→C(x,y,z),
u=∫[(0,0,0),(x,0,0)] yzdx+∫[(x,0,0),(x,y,0)] xzdy+∫[(x,y,0),(x,y,z)]xydz+C
=0+0+xyz+C
=xyz+C.
3、AB方程为:y=-x+1,
y’=-1,ds=√(1+y’^2)dx=√2dx,
∫[AB](x+y)ds=∫[0,1]*1*√2dx=√2(1-0)=√2.
3\、 z=1-x-y,
∂z/∂x=-1, ∂z/∂y=-1,
√[(1+(-1)^2+(-1)^2)=√3,
dS=√[1+(∂z/∂x)^2++( ∂z/∂y)^2]dxdy= √3dxdy
∫[Σ]∫(x+y)dS=∫[D]∫(x+y)√3dxdy
=√3∫[0,1]dx∫[0,-x+1](xy+y^2/2)
=√3∫[0,1] (1/2-x^2/2)dx
=√3(x/2-x^3/6)[0,1]
=√3/3.