跪解一道利用对称性计算二重积分的设D为x方+y方≤1,D1为x方+y方≤1,x≥0,则使∫∫D f(x,y)dxdy=2∫∫D1 f(x,y)dxdy成立的充分条件是?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 07:05:24
![跪解一道利用对称性计算二重积分的设D为x方+y方≤1,D1为x方+y方≤1,x≥0,则使∫∫D f(x,y)dxdy=2∫∫D1 f(x,y)dxdy成立的充分条件是?](/uploads/image/z/13769986-58-6.jpg?t=%E8%B7%AA%E8%A7%A3%E4%B8%80%E9%81%93%E5%88%A9%E7%94%A8%E5%AF%B9%E7%A7%B0%E6%80%A7%E8%AE%A1%E7%AE%97%E4%BA%8C%E9%87%8D%E7%A7%AF%E5%88%86%E7%9A%84%E8%AE%BED%E4%B8%BAx%E6%96%B9%2By%E6%96%B9%E2%89%A41%2CD1%E4%B8%BAx%E6%96%B9%2By%E6%96%B9%E2%89%A41%2Cx%E2%89%A50%2C%E5%88%99%E4%BD%BF%E2%88%AB%E2%88%ABD+f%28x%2Cy%29dxdy%3D2%E2%88%AB%E2%88%ABD1+f%28x%2Cy%29dxdy%E6%88%90%E7%AB%8B%E7%9A%84%E5%85%85%E5%88%86%E6%9D%A1%E4%BB%B6%E6%98%AF%3F)
跪解一道利用对称性计算二重积分的设D为x方+y方≤1,D1为x方+y方≤1,x≥0,则使∫∫D f(x,y)dxdy=2∫∫D1 f(x,y)dxdy成立的充分条件是?
跪解一道利用对称性计算二重积分的
设D为x方+y方≤1,D1为x方+y方≤1,x≥0,则使∫∫D f(x,y)dxdy=2∫∫D1 f(x,y)dxdy成立的充分条件是?
跪解一道利用对称性计算二重积分的设D为x方+y方≤1,D1为x方+y方≤1,x≥0,则使∫∫D f(x,y)dxdy=2∫∫D1 f(x,y)dxdy成立的充分条件是?
f(x,y)=f(-x,y)
其实你写f(x,y)=0也是充分条件啊.