二重積分積分區(qū)域的對稱性

1、情形一：情形一：積分區(qū)域積分區(qū)域關(guān)于坐標(biāo)軸對稱關(guān)于坐標(biāo)軸對稱D定理定理4設(shè)二元函數(shù)在平面區(qū)域連續(xù)，且關(guān)于軸對稱，則()fxyDDx1)當(dāng)（即是關(guān)于的奇函數(shù)）時(shí)，有()()fxyfxy???()fxyy.()0Dfxydxdy???2)當(dāng)（即是關(guān)于的偶函數(shù)）時(shí)，有()()fxyfxy??()fxyy.1()2()DDfxydxdyfxydxdy?????其中是由軸分割所得到的一半?yún)^(qū)域。1DxD例5計(jì)算，其中為由與圍成的區(qū)域。3()DIxy

2、ydxdy????D22yx?2x?解：如圖所示，積分區(qū)域關(guān)于軸對稱，且Dx3()()()fxyxyyfxy??????即是關(guān)于的奇函數(shù)，由定理1有()fxyy.3()0Dfxyydxdy????類似地，有：定理定理5設(shè)二元函數(shù)在平面區(qū)域連續(xù)，且關(guān)于軸對稱，則()fxyDDy22()()().()0()().DDfxydxdyfxyfxyfxydxdyfxyfxy??????????????當(dāng)當(dāng)其中是由軸分割所得到的一半?yún)^(qū)域。2DyD例

3、6計(jì)算其中為由2DIxydxdy???D所圍。22220yxyxy?????及解：如圖所示，關(guān)于軸對稱，并且Dy，即被積分函數(shù)是關(guān)于2()()fxyxyfxy???軸的偶函數(shù)，由對稱性定理結(jié)論有：x解:如圖所示區(qū)域關(guān)于原點(diǎn)對稱對于被積函數(shù)有D33()fxyxy??有定理7得3333()()()()()fxyxyxyfxy???????????.33()0Dxydxdy????情形三、情形三、積分區(qū)域積分區(qū)域關(guān)于直線關(guān)于直線對稱對稱Dyx

4、??定理定理8設(shè)二元函數(shù)在平面區(qū)域連續(xù)且關(guān)于直線對()fxyD12DDD??12DDyx?稱則1)()()DDfxydxdyfyxdxdy?????.12()()DDfxydxdyfxydxdy?????2)當(dāng)時(shí)有.()()fyxfxy??()0Dfxydxdy???3)當(dāng)時(shí)有.()()fyxfxy?1()2()DDfxydxdyfxydxdy?????例9求為所圍.2222()DxyIdxdyab????D222xyR??解:積分區(qū)域

5、關(guān)于直線對稱由定理8得Dyx?22222222()()DDxyyxdxdydxdyabab???????故2222()DxyIdxdyab????222222221[()()]2DDxyyxdxdydxdyabab????????2222111()()2Dxydxdyab?????222200111()2Rdrrdrab??????.42211()4Rab???類似地，可得：定理定理9設(shè)二元函數(shù)在平面區(qū)域連續(xù)且關(guān)于直線對稱則()fxyD