计算下列对坐标的曲线积分:
( 1 ) ∮ L x y d x , (1)\oint_{L}xydx, (1)∮Lxydx,其中 L L L为圆周 ( x − a ) 2 + y 2 = a 2 ( a > 0 ) (x-a)^2+y^2=a^2(a>0) (x−a)2+y2=a2(a>0)及x轴所围成的在第一象限内的区域的整个边界(按逆时针方向绕行);
( 2 ) ∮ Γ d x − d y + y d z , (2)\oint_{\Gamma}dx-dy+ydz, (2)∮Γdx−dy+ydz,
其中 Γ \Gamma Γ为有向折线 A B C A , ABCA, ABCA,这里的 A , B , C A,B,C A,B,C依次为点 ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) ; (1,0,0),(0,1,0),(0,0,1); (1,0,0),(0,1,0),(0,0,1);
( 3 ) ∫ L ( x + y ) d x + ( y − x ) d y , (3)\int_{L}(x+y)dx+(y-x)dy, (3)∫L(x+y)dx+(y−x)dy,其中 L L L是曲线 x = 2 t 2 + t + 1 , y = t 2 + 1 , x=2t^2+t+1,y=t^2+1, x=2t2+t+1,y=t2+1,从点 ( 1 , 1 ) (1,1) (1,1)到点 ( 4 , 2 ) (4,2) (4,2)的一段弧.
解:
( 1 ) L (1)L (1)L由 L 1 ( 圆 弧 ) L_1(圆弧) L1(圆弧)和 L 2 ( 直 线 段 ) L_2(直线段) L2(直线段)组成. L 1 L_1 L1为有向半圆弧:
{ x = a + a c o s t , y = a s i n t , \left\{\begin{aligned} &x=a+acost,\\ &y=asint, \end{aligned}\right. { x=a+acost,y=asint,
t t t从0变到 π ; \pi; π;
L 2 L_2 L2为有向线段 y = 0 , x y=0,x y=0,x从 0 0 0变到 2 a . 2a. 2a.于是
∮ L x y d x = ∫ L 1 x y d x + ∫ L 2 x y d x = ∫ 0 π a ( 1 + c o s t ) ⋅ a s i n t ⋅ ( − a s i n t ) d t + 0 = − a 3 ( ∫ 0 π s i n 2 t d t + ∫ 0 π s i n 2 t c o s t d t ) = − a 3 ( π 2 + 0 ) = − π 2 a 3 . \oint_{L}xydx=\int_{L_1}xydx+\int_{L_2}xydx\\ =\int_{0}^{\pi}a(1+cost) \cdot asint \cdot (-asint)dt+0\\ =-a^3(\int_{0}^{\pi}sin^2tdt+\int_{0}^{\pi}sin^2tcostdt)\\ =-a^3(\frac{\pi}{2}+0)=-\frac{\pi}{2}a^3.\\ ∮Lxydx=∫L1xydx+∫L2xydx=∫0πa(1+cost)⋅asint⋅(−asint)dt+0=−a3(∫0πsin2tdt+∫0πsin2tcostdt)=−a3(2π+0)=−2πa3.
( 2 ) (2) (2) Γ \Gamma Γ由有向线段AB,BC,CA依次连接而成,其中
A B : x = 1 − t , y = t , z = 0 , t 从 0 变 到 1 ; B C : x = 0 , y = 1 − t , z = t , t 从 0 变 到 1 ; C A : x = t , y = 0 , z = 1 − t , t 从 0 变 到 1 : AB:x=1-t,y=t,z=0,t从0变到1;\\ BC:x=0,y=1-t,z=t,t从0变到1;\\ CA:x=t,y=0,z=1-t,t从0变到1:\\ AB:x=1−t,y=t,z=0,t从0变到1;BC:x=0,y=1−t,z=t,t从0变到1;CA:x=t,y=0,z=1−t,t从0变到1:
∫ A B d x − d y + y d z = ∫ 0 1 [ ( − 1 ) − 1 + 0 ] d t = − 2 , ∫ B C d x − d y + y d z = ∫ 0 1 [ 0 − ( − 1 ) + ( 1 − t ) ⋅ 1 ] d t = ∫ 0 1 ( 2 − t ) d t = 3 2 , ∫ C A d x − d y + y d z = ∫ 0 1 ( 1 − 0 + 0 ) d t = 1 , \int_{AB}dx-dy+ydz=\int_{0}^{1}[(-1)-1+0]dt=-2,\\ \int_{BC}dx-dy+ydz=\int_{0}^{1}[0-(-1)+(1-t) \cdot 1]dt=\int_{0}^{1}(2-t)dt=\frac{3}{2},\\ \int_{CA}dx-dy+ydz=\int_{0}^{1}(1-0+0)dt=1, ∫ABdx−dy+ydz=∫01[(−1)−1+0]dt=−2,∫BCdx−dy+ydz=∫01[0−(−1)+(1−t)⋅1]dt=∫01(2−t)dt=23,∫CAdx−dy+ydz=∫01(1−0+0)dt=1,
因此
∮ Γ d x − d y + y d z = − 2 + 3 2 + 1 = 1 2 . \oint_{\Gamma}dx-dy+ydz=-2+\frac{3}{2}+1=\frac{1}{2}. ∮Γdx−dy+ydz=−2+23+1=21.
( 3 ) (3) (3)由 { 2 t 2 + t + 1 = 1 , t 2 + 1 = 1 \left\{\begin{aligned} &2t^2+t+1=1,\\ &t^2+1=1\\ \end{aligned} \right. { 2t2+t+1=1,t2+1=1
可得 t = 0 ; t=0; t=0;
由 { 2 t 2 + t + 1 = 4 , t 2 + 1 = 2 \left\{\begin{aligned} &2t^2+t+1=4,\\ &t^2+1=2\\ \end{aligned} \right. { 2t2+t+1=4,t2+1=2
可得 t = 1 ; t=1; t=1;
因此
∫ L ( x + y ) d x + ( y − x ) d y = ∫ 0 1 [ 2 t 2 + t + 1 + t 2 + 1 ) ⋅ ( 4 t + 1 ) + ( t 2 + 1 − 2 t 2 − t − 1 ) ⋅ 2 t ] d t = ∫ 0 1 ( 10 t 3 + 5 t 2 + 9 t + 2 ) d t = 32 3 . \int_{L}(x+y)dx+(y-x)dy=\int_{0}^{1}[2t^2+t+1+t^2+1) \cdot (4t+1)+(t^2+1-2t^2-t-1) \cdot 2t]dt \\ =\int_{0}^{1}(10t^3+5t^2+9t+2)dt=\frac{32}{3}. ∫L(x+y)dx+(y−x)dy=∫01[2t2+t+1+t2+1)⋅(4t+1)+(t2+1−2t2−t−1)⋅2t]dt=∫01(10t3+5t2+9t+2)dt=332.