Question

帮我解一道数学题！谢谢。很急！

计算曲线积分∫D(5x^2-2y^2+x^2*e^3y)dx+(x^3*e^3y-4xy+5y^2)dy,其中D为上半圆周(x-1)^2+y^2=1,y>=0,沿逆时针方向.

我需要此题详细解答过程。谢谢！

Answer 1

计算曲线积分∫D(5x²-2y²+x²e³y)dx+(x³e³y-4xy+5y²)dy,其中D为上半圆周(x-1)²+y²=1,y≧0,沿逆时针方向.
解：将积分曲线D的方程改写成参数形式：D：x=1+cost，y=sint，0≦t≦π
dx/dt=-sint，dy/dt=cost，于是：
(D)∫(5x²-2y²+x²e³y)dx+(x³e³y-4xy+5y²)dy
=[0，π]∫{[5(1+cost)²-2sin²t+(1+cost)²e³sint](-sint)+[(1+cost)³e³sint-4(1+cost)sint+5sin²t]cost}dt
=[0，π]{∫5(1+cost)²d(1+cost)+2∫sin³tdt-∫e³(1+cost)²(1-cos²t)dt-∫e³(1+cost)³costd(cost)
+4∫(1+cost)costd(cost)+5∫sin²td(sint)}
={(5/3)(1+cost)³-2[cost-(cos³t)/3]-e³∫[1+2cost-2cos³t-(cost)^4]dt-e³∫[cost+3cos²t+3cos³t+
+(cost)^4]d(cost)+4∫(cost+cos²t)d(cost)+(5/3)sin³t}[0，π]
={(5/3)(1+cost)³-2[cost-(cos³t)/3]-e³[t+2sint-2(sint-(sin³t)/3)-(cos³tsint)/4-(3/4)(t/2+(1/4)sin2t)]
-e³[(1/2)cos²t+cos³t+(3/4)(cost)^4+(1/5)(cost)^5]+4[(1/2)cos²t+(1/3)cos³t+(5/3)sin³t}[0，π]
={-2(-1+1/3)-e³[π-(3/4)(π/2)]-e³[(1/2)-1+(3/4)-(1/5)]+4[(1/2)-(1/3)}-{(40/3)-2(1-1/3)-e³×0
-e³[(1/2)+1+(3/4)+(1/5)]+4[(1/2)+(1/3)+0}
={4/3-(5/8)e³π+(1/20)e³+2/3}-{(40/3)-(4/3)-(49/20)e³+20/6}
=-40/3-(5/8)e³π-(12/5)e³=-40/3-(5π/8+12/5)e³
[我不能保证最后结果一定正确，因为计算太复杂！但方法保证是正确的！]

Answer 2

因为AD是∠BAC的平分线
所以∠BAD=∠CAD
又因为∠B=∠EAC
所以∠BAD+∠B=∠CAD+∠EAC
即∠ADE=DAE
所以AE=DE
又因为EF垂直AD
所以EF平分∠AEB(等腰三角形三线合一)