可化為2/pi *(sina+1)cos^2a da=2/pi *sina*cos^2a da + 2/pi *cos^2a da
前一部分是奇函式,所以積分為0,後部分結果是1
2樓:匿名使用者
∫[0,2 ] 2(y/π)√(2y-y^2)dy)=(2/π)∫[0.2]y√(2y-y^2)dy
=(2/π)∫[0,2]y√(2y-y^2)dy
=(2/π)∫[0,2]y√[1-(y-1)^2]dy
y-1=sinu y=0 u=-π/2 y=2,u=π/2
=(2/π)∫[-π/2,π/2] (1+sinu)cosudsinu
=(2/π)∫[-π/2,π/2] (1+sinu)cosu^2du
=(2/π) ∫[-π/2,π/2](1+sinu)(1-sinu^2)du
=(2/π) ∫[-π/2,π/2](1+sinu-sinu^2-sinu^3)du
=(2/π)[1-cosu+(1/2)(cos2u-1)+cosu-cosu^3/3] |[-π/2,π/2]
=(2/π)(π/2)=1
3樓:匿名使用者
∫(0->2) (2y/π)√(2y-y²) dy = (2/π)l
l = ∫(0->2) y√(2y-y²) dy
= ∫(0->2) y√[1-(y-1)²] dy
令y = 1 + sinz,dy = cosz dz
當y = 0,z = -π/2,當y = 2,z = π/2
l = ∫(-π/2->π/2) (1+sinz)cos²z dz
= ∫(-π/2->π/2) cos²z dz + ∫(-π/2->π/2) sinz*cos²z dz
= (1/2)∫(-π/2->π/2) (1+cos2z) dz - ∫(-π/2->π/2) cos²z d(cosz)
= (1/2)[z+(1/2)sin(2z)] - (1/3)cos³z
= (1/2)(π/2 - (-π/2)) - 0
= π/2
∴∫(0->2) (2y/π)√(2y-y²) dy = (2/π)l = (2/π)(π/2)= 1
∫(x^2-y)dx-(x+sin^2y)dy,其中l是圓周y=根號下2x-x^2上由點(0,0)到(1,1)上一段弧。不用格林公式
30
4樓:世間狂傲r律己
p(x,y) = (x^2-y) , 偏p/偏y = -1
q(x,y) = -(x+(siny)^2) , 偏q/偏x = -1
偏q/偏x - 偏p/偏y = (-1) - (-1) = 0
由格林公式, 知,
原式 = (l1)∫l(x^2-y)dx-(x+(siny)^2)dy,
其中 l1 : a(0,0) --> b(1,0) ---> c(0,1)
原式 = (0,1)∫l(x^2-0)dx - (0,1)∫[1+(siny)^2]dy
= 1/3 - (0,1)∫[3/2 - (1/2)* cos(2y)]dy
= 1/3 - (0,1)∫[3/2 - (1/2)* cos(2y)]dy
= - 7/6 + (1/4)*sin2
5樓:茹翊神諭者
簡單計算一下即可,答案如圖所示
求∫(1,2)(-2y/π)*cos((πy)/2)dy,求詳細過程
6樓:匿名使用者
∫(1,2) (-2y/π)*cos(πy/2)*dy=∫(1,2) (-4y/π^2)*d[sin(πy/2)]=(-4y/π^2)*sin(πy/2)|(1,2)-∫(1,2) sin(πy/2)*d(-4y/π^2)
=4/π^2+∫(1,2) (4/π^2)*sin(πy/2) dy=4/π^2-(8/π^3)*cos(πy/2)|(1,2)=4/π^2-(-8/π^3)
=4/π^2+8/π^3
7樓:匿名使用者
∫<1,2>(-2y/π)*cos(πy/2)dy=-4y/π^2*sin(πy/2)|<1,2>+4/π^2∫<1,2>sin(πy/2)dy
=4/π^2-8/π^3*cos(πy/2)|<1,2>=4/π^2+8/π^3.
求∫l(x^2-y)dx-(x+sin^2y)dy,l是y=根號下1-x^2以a(-1,0)到b(1,0)
8樓:匿名使用者
在圓弧l下補一條線n:y = 0,反向。
∮(l+n) (x² - y)dx - (x + sin²y)dy= - ∫∫d [ ∂/∂x (- x - sin²y) - ∂/∂y (x² - y) ] dxdy
= - ∫∫d [ - 1 - (- 1) ] dxdy= 0∫n (x² - y)dx - (x + sin²y)dy= ∫(1→- 1) x² dx
= - 2∫(0→1) x² dx
= - 2/3
因此∫l (x² - y)dx - (x + sin²y)dy = 0 - (- 2/3) = 2/3
格林公式,求∫l(x^2+y^2)dx+(1+2y)dy,l沿y=√2x-x^2由(0,0)到(2,0)
9樓:匿名使用者
py=1 qx=-1
qx-py=-2
由格林公式:
∫+l(x+y)dx-(x-y)dy
=∫∫(-2)dxdy
=-2πab
10樓:匿名使用者
一定要用格林公式還是什麼?