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Cos(x+y)的绝对值且区域由y=x和y=0.x=兀/2构成的二...
2020-11-21

解 记D={(x,y):0≤y≤x,0≤x≤π2},D1={(x,y)∈D:x+y≤π2},D2={(x,y)∈D:x+y>π2},则?Dcos(x+y)dxdy=?D1[-cos(x+y) ]dxdy+?D2|cos(x+y)|dxdy=∫π40dy∫π2?yycos(x+y)dx+∫π2π4dx∫xπ2?x[?cos(x+y)]dy=∫π40(1?sin2y)dy+∫π2π4(1?sin2x)dx=∫

解:∫(0,π)dy∫(0,y)cos(x+y)dx=∫(0,π)(sin2y-siny)dy=(-1/2)*cos2y+cosy | (0,π)=0+(-1-1)=-2

∫∫_d cos(x + y) dσ= ∫(0→π) dy ∫(0→y) cos(x + y) dx= ∫(0→π) dy ∫(0→y) cos(x + y) d(x + y)= ∫(0→π) sin(x + y) |(0→y) dy= ∫(0→π) [sin(2y) - sin(y)] dy= cos(y) - (1/2)cos(2y) |(0→π)= [cos(π) - (1/2)cos(2π)] - [cos(0) - (1/2)cos(0)]= - 2

x和y可互换,因此原式 = 2∫∫丨 cos(x+y) 丨dxdy 区域为d0: 0<=x<=π/2 0<=y<=x = 2[∫∫<积分区域d1> cos(x+y) dxdy + ∫∫<积分区域d2> -cos(x+y) dxdy ] d1: 0<=y<=π/4, y<=x<=π/2 -y d2: 0<=x<=π/4, x<=y<=π/2 -x d0=d1∪d2 则原式= 2[∫∫<积分区域d1>

∫∫_D cos(x + y) dσ= ∫(0→zhidaoπ) dy ∫(0→y) cos(x + y) dx= ∫(0→π) dy ∫(0→y) cos(x + y) d(x + y)= ∫(0→π) sin(x + y) |(0→y) dy= ∫(0→π) [sin(2y) - sin(y)] dy= cos(y) - (1/2)cos(2y) |(0→π)= [cos(π) - (1/2)cos(2π)] - [cos(0) - (1/2)cos(0)]= - 2

∫∫_[D] cos(x + y) dxdy= ∫[0→π] dy ∫[0→y] cos(x + y) dx= ∫[0→π] {sin(x + y) |[0→y]} dy= ∫[0→π] {sin(y + y) - sin(0 + y)} dy= ∫[0→π] (sin2y - siny) dy= {(- 1/2)cos2y + cosy} |[0→π]= {cos(π) - (1/2)cos(2π)} - {cos(0) - (1/2)cos(0)}= - 2

先去绝对值符号,由题中给的D可以知道0≤x+y≤π,所以用x+y=二分之π将积分D分成两部分,去绝对值之后I=∫∫D1-∫∫D2=∫0→π/2dx∫0→π/2-x cos(x+y)dy-∫0→π/2dx∫π/2-x→π/2 cos(x+y)dy=(π/2-1)-(1-π/2)=π-2

解:∵a=[0,π]*[0,π] ∴0≤x+y≤2π ∵当0≤x+y≤π/2时,cos(x+y)≥0 当π/2≤x+y≤3π/2时,cos(x+y)≤0 当3π/2≤x+y≤2π时,cos(x+y)≥0 ∴∫∫|cos(x+y)|dxdy=∫dx[∫cos(x+y)dy-∫cos(x+y)dy] +∫dx[-∫cos(x+y)dy+∫cos(x+y)dy] =∫[(sin(π/2)-sinx)-(sin(π+x)-sin(π/2))]dx +∫[(sin(π+x)-sin(3π/2))-(sin(3π/2)-sinx)]dx =2∫dx+2∫dx =2*(π/2)+2*(π-π/2) =2π.

这两道题型都差不多,都将二重积分化为极坐标形式的累次积分. 1) D:x^2+y^2<=x+y ,所以 D:(x-1/2)^2+(y-1/2)^2<=1/2. 画出(图1-1),θ: [-π/4, 5π/4] 设 ρ(θ)=ρ*cosθ 则积分区域D用不等式组表示为: D: -π/4<= θ<=5π/4 (√2-1)/2*cosθ<= ρ<= (1+

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