∵積分割槽域D關於x，y的對稱性∴?Dxsin(πx2+y2)x+ydxdy=?Dysin(πx2+y2)x+ydxdy因此有：?Dxsin(πx2+y2)x+ydxdy=12[?Dxsin(πx2+y2)x+ydxdy+?Dysin(πx2+y2)x+ydxdy]=12?D(x+y)sin(πx2+y2)x+ydxdy=12?Dsin（πx2+y2）dxdy令x=rcosθ，y=rsinθ r∈[1，2]，θ?[0，π2]；於是原積分可化為：?Dxsin(πx2+y2)x+ydxdy=12?Dsin（πx2+y2）dxdy=12∫π20dθ∫21rsinπrdr=π4∫21rsinπrdr=14∫21rsinπrdπr=14×（-∫21rdcosπr）=14×（-rcosπr|21+∫21cosπrdr）=14×（-3+∫21cosπrdr）=-34．故本題答案為：-34．
∵積分割槽域D關於x，y的對稱性∴?Dxsin(πx2+y2)x+ydxdy=?Dysin(πx2+y2)x+ydxdy因此有：?Dxsin(πx2+y2)x+ydxdy=12[?Dxsin(πx2+y2)x+ydxdy+?Dysin(πx2+y2)x+ydxdy]=12?D(x+y)sin(πx2+y2)x+ydxdy=12?Dsin（πx2+y2）dxdy令x=rcosθ，y=rsinθ r∈[1，2]，θ?[0，π2]；於是原積分可化為：?Dxsin(πx2+y2)x+ydxdy=12?Dsin（πx2+y2）dxdy=12∫π20dθ∫21rsinπrdr=π4∫21rsinπrdr=14∫21rsinπrdπr=14×（-∫21rdcosπr）=14×（-rcosπr|21+∫21cosπrdr）=14×（-3+∫21cosπrdr）=-34．故本題答案為：-34．