例1。有界区域,求围成的平面及是由曲线其中满足设连续函数),(),()cos(),(),(2yxfxyxyDdxdyyxfxyxexyxfyxfDy解332),()cos(),(),(102aedxdyadyxedxdxdyyxfaaxyxexyxfadxdyyxfDxxyDyD239ea239)cos(),(exyxexyxfy例2解.10,11:.2yxDdxyD其中计算先去掉绝对值符号,如图dxydyxdxyDDDD321)()(2221211021122)()(xxdyxydxdyyxdx.15111D2D3D例3解).所围的面积(取圆外部和圆是由心脏线其中计算ararDdyxD)cos1(.22)cos1(2222aaDrdrrddyx2233]1)cos1[(31da).2922(3a)0(.),(22202adyyxfdxIaxxaxa更换积分次序例4解,22,20:2axyxaxaxD,,321三部分及分成将积分区域DDDD2D1D3D;0,2:2221ayyaaxayD;2,22:22ayaaxayD;0,2:223ayaxyaaD.),(),(),(2022220222222ayaaaaayaayaaayadxyxfdydxyxfdydxyxfdyI故1,222zyxdvez:其中求例5解法一11zDzzdxdydzedve2221:zyxDz2)1(2)1(210112dzzedzzezz22424)1(210110120101222tdterdrerdredttrrr令解法二12dvedvezzdxdyedzedxdyDDyxyxz)1(22222211001:2221zzyx及1:22yxD12dvedvezz2]cos2cos2cos1[sin4sin4sin2sin22221032cos0210cos210cos2002cosdedrreddrreddddrdrerrr原式01:2221zzyx及解法三1)(222222222222czbyaxdvczbyax:其中求,1:'cos,sinsin,cossinrcrzbryarx令zzzyyyxxxJrrrsin0sincoscossinsincossinsinsinsincoscoscossin2abcrcrcbrbrbarara例6解法一ddrdabcrrdvczbyaxrsin)(212222222abcdrabcrdd20010454sinabcaxaxbcdxaxbcaxdxaxbcaxaaaa154]53[2)1(2)1(04523022222222)1(,1:22222222axbcSaxczbyDxx...