计算二重积分∫∫(x^2-y^2)^(1/2)dxdy,D是以(0,0),(1,-1),(1,1)为...答:积分区域D关于x轴对称,原式=2∫∫[D1](x^2-y^2)^(1/2)dxdy, D1为y=x,x=1,y=0围成的区域 =2∫[0->1]∫[0->x] (x^2-y^2)^(1/2)dydx 换元y=xcost, t∈[-π/2,0]=2∫[0->1]∫[-π/2->0] -xsint(x^2-y^2)^(1/2)dtdx =2∫[0->1]∫[-π/2-...
二重积分计算,换元法计算∫∫(x-y)²sin²(x+y)d x dy,积分区域为...答:解:∵由(π,0),(2π,π),(π,2π),(0,π)四点所确定平行四边形,是由x-y=-π,x-y=π,x+y=π,x+y=3π 四条直线所围成 ∴做变换u=x-y,v=x+y,则-π≤u≤π,π≤v≤3π ==>x=(u+v)/2,y=(v-u)/2 ==>dxdy=│J│dudv=(1/4)dudv (│J│是雅可比行列式...