大一高数计算二重积分 高等数学，计算二重积分？
\u5927\u4e00\u9ad8\u6570\u8ba1\u7b97\u4e8c\u91cd\u79ef\u5206\u89e3\uff1a1\u5927\u9898(1)\u5c0f\u9898\uff0cD={(x,y)\u4e280\u2264x\u22641\uff0cx\u2264y\u22642x}\u3002\u2234\u539f\u5f0f=\u222b(0,1)dx\u222b(x,2x)(x²-y²)dy=\u222b(0,1)(-4x³/3)dx=-1/3\u3002(2)\u5c0f\u9898\uff0cD={(x,y)\u4e28-1\u2264x\u22640\uff0c-x-1\u2264y\u2264x+1}\u222a{(x,y)\u4e280\u2264x\u22641\uff0cx-1\u2264y\u22641-x}\u3002\u2234\u539f\u5f0f=\u222b(-1,0)dx\u222b(-x-1,x+1)(y-x)dy+\u222b(0,1)dx\u222b(x-1,1-x)(y+x)dy=\u2026=4\u222b(0,1)x(1-x)dx= 2/3\u3002(3)\u5c0f\u9898\uff0cD={(x,y)\u4e280\u2264x\u22641\uff0c0\u2264y\u2264x}\u3002\u2234\u539f\u5f0f=\u222b(0,1)xdx\u222b(0,x)sin(y/x)dy=(1-cos1) \u222b(0,1)xdx=(1-cos1)/2\u3002(4)\u5c0f\u9898\uff0cD={(x,y)\u4e280\u2264x\u2264y\uff0c\u03c0/2\u2264y\u2264\u03c0}\u3002\u2234\u539f\u5f0f=\u222b(\u03c0/2,\u03c0)dy\u222b(0,y) sinydx/y=\u222b(\u03c0/2,\u03c0)siny=1\u3002(5)\u5c0f\u9898\uff0cD={(x,y)\u4e28-1\u2264x\u22641\uff0cx²\u2264y\u22641}\u222a{(x,y)\u4e28-1\u2264x\u22641\uff0c0\u2264y\u2264x²}\u3002\u2234\u539f\u5f0f=\u222b(-1,1)dx\u222b(x²,1)(y-x²)dy+\u222b(-1,1)dx\u222b(0,x²)(x²-y)dy=\u222b(-1,1)(1/2-x²+x^4) dx=11/15\u3002
2\u5927\u9898(1)\u5c0f\u9898\uff0c \u8bbex=\u03c1cos\u03b8\uff0cy=\u03c1sin\u03b8\u3002\u22340\u2264\u03b8\u22642\u03c0\uff0c1\u2264\u03c1\u22642\u3002 \u2234\u539f\u5f0f=\u222b(0,2\u03c0)d\u03b8\u222b(1,2) \u03c1d\u03c1 /(1+\u03c1²)=\u03c0ln(5/2)\u3002(2)\u5c0f\u9898\uff0c \u8bbex=\u03c1cos\u03b8\uff0cy=\u03c1sin\u03b8\u3002\u22340\u2264\u03b8\u22642\u03c0\uff0c0\u2264\u03c1\u2264a\u3002 \u2234\u539f\u5f0f=\u222b(0,2\u03c0)\u4e28cos\u03b8sin\u03b8\u4e28d\u03b8\u222b(0,a)\u03c1³d\u03c1=(a^4/8)\u222b(0,2\u03c0)\u4e28sin2\u03b8\u4e28d\u03b8=(a^4)/2\u3002(3)\u5c0f\u9898\uff0c \u8bbex=\u03c1cos\u03b8\uff0cy=\u03c1sin\u03b8\u3002\u22340\u2264\u03b8\u2264\u03c0/4\uff0c1\u2264\u03c1\u22642\u3002 \u2234\u539f\u5f0f=\u222b(0,\u03c0/4)arctan(tan\u03b8)d\u03b8\u222b(1,2)\u03c1d\u03c1=(3/2)\u222b(0,\u03c0/4) arctan(tan\u03b8)d\u03b8\u3002\u4ee4t=tan\u03b8\uff0c\u539f\u5f0f=(3/2)\u222b(0,1)arctantdt/(1+t²)=3\u03c0²/16\u3002
\u4f9b\u53c2\u8003\u3002
如图
x^2+y^2 = x+y, 化为极坐标， r = cost+sint
I = ∫<-π/4, 3π/4>dt ∫<0, cost+sint>r(cost+sint) rdr
= ∫<-π/4, 3π/4>(cost+sint)dt ∫<0, cost+sint>r^2 dr
= (1/3)∫<-π/4, 3π/4>(cost+sint)^4dt = (1/3)∫<-π/4, 3π/4>(1+sin2t)^2dt
= (1/3)∫<-π/4, 3π/4>[1+2sin2t+(sin2t)^2]dt
= (1/3)∫<-π/4, 3π/4>[3/2+2sin2t+(1/2)cos4t]dt
= (1/3)[3t/2 - cos2t + (1/8)sin4t]<-π/4, 3π/4> = π
大学不学高数是我最开心的
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