【正文】 x2??xdy?0 ?y2?222?1，且此區(qū)域面積為?，則2?x?y?3當2?x2?y2?3??xdy???22且此區(qū)域面積為?,則3?x?y?4當3?x?y?4??xdy???0?????0故 x?y?43．試用二重積分的定義證明: 222??2xdy?.1(1) ??d??SDD(其中SD為D之面積)D(2) ??kf(x,y)d??k??f(x,y)d?D(k為常數(shù))n?0i.證 (1) 由二重積分的定義,有?f(???f(x,y)d??lim?Di?1,?i)??i??f(x,y)?1則當時,上式變?yōu)镈n?od??lim???i?limSD?SD??0i?1n??0. (2) 由二重積分的定義,有 ?kf(??)????kf(x,y)d??lim?i,iDi?1ni?limk?f(?i,?i)??i??oi?1n?klim?f(?　i,?i)??i??0i?14．根據(jù)二重積分的性質,比較下列積分的大小. D ?k??f(x,y)d?.2(1) ???x?y?d?D與??(x?y)d?D33,其中D由x軸、y軸及直線x?y?1圍成。 (2) ???x?y?d?D2與??(x?y)d?D2(x?2)? ，其中D由圓(y?1)2?2圍成.解 (1) 積分區(qū)域D如圖9－1 所示.因在所圍區(qū)域內有0?x?y?1,23(x?y)?(x?y)所以2D故 D. 圖9－1(2) 積分區(qū)域D如圖9－2 所示. ???x?y?d????(x?y)d?3因圓(x?2)?(y?1)?2的參數(shù)方程為??x?2?????y?1?? 22x?y?3??cos?)?3?2sin(??)4 圖9－2 則而(x?y)min?3?2?1,且x?y?1,于是(x?y)2?(x?y)3D故 D5．利用二重積分的性質,估計下列積分的值. 3x?yd??(x?y)d?.??????2?(1) I???xy(x?y)d?D， D:0?x?1,0?y?12(2) I???sin2xsin2yd?D， D:0?x??,0?y??(3) I???(x?y?1)d?D， D:0?x?1,0?y?222D:x?y?4 ， (4) I???(x2?4y2?9)d?D?0?x?1?0?xy?1,??0?y?1解 (1) 因?則?0?x?y?2故 0???0d??I???2d??2??d??2SD?2.DDD?0?x???0?sinx?1,??0?y??(2) 因?則?0?siny?1于是 0?sinxsiny?1故 220???0d??I???d??SD??2.DD?0?x?1?（3）因?0?y?2，則1?x?y?1?4故 ??d????(x?y?1)d??4??d?DDD即 2?I?8.2222 (4) 因0?x?y?4,則9?x?4y?9?4(x?y)?9?25 22于是 9SD???9d??I???25d??25SDDD而 SD??r?4?故 36??I?100?.2習題 9－2：(1) ??(x2?y2)d?,D其中D是矩形區(qū)域：x?1,y?1。其中D由直線y?y?x與y?2x所圍成。 (2) ??(x2?y2?x)d?,D(3) ??xy2d?,D2其中D由拋物線y?x和直線y?x所圍成。(4) ?dy1221解 (1)9－3 所示. 22(x?y)d??dx(x?y)dy????D?1?1211x.28??(2x2?)dx?.?133 圖9－3 13(2)積分區(qū)域D如圖9－4所示.??(x2?y2?x)d???2dy?y01(x2?y2?x)dxD2y??2(1933024y?8y2)dy?136 圖9－4(3)積分區(qū)域D如圖9－5所示.??xy2d??1x21D?0dx?13xx2xydy??0(x?3yx2)dx??1(1x4?1x71033)dx?40 圖 9－5(4)積分區(qū)域D如圖9－6所示.?2dy111x??0dx?x2?1sinx1xy??10xsinxdx?sin1?cos1. 圖9－62．積分區(qū)域D??(x,y)a?x?b,c?y?d?，且被積函數(shù)為f(x)?g(y),求證：??f(x)?g(y)dxdy?bdy)dD?afx()dx?cgy(.證 積分區(qū)域D如圖9－7所示.??f(x)g(y)dxdy?bdD?adx?cf(x)g(y)dy??bf(x)dx??dg(y)dy?bc?af(x)dx??bf(x)dx??dacg(y)dy? 圖9－73．設f(x,y)在D上連續(xù)且D由y?x、y?a與x?b（b?a）圍成，xbb求證：?badx?af(x,y)dy??ady?yf(x,y)dx.證 積分區(qū)域D如圖9－8 所示.交換等式左邊二次積分的積分順序有?bxadx?f(x,y)dy??bdy?baayf(x,y)dx 圖9－8I?4．下列條件下，將??f(x,y)d?D按不同積分順序化為二次積分：(1) D由y?x與y2?4x所圍成；(2) D由x軸與222半圓周x?y?r?y?0?所圍成.解 (1) 由y2?4x和y?x,得交點為(0,0),(4,4).4積分區(qū)域D如圖9－9 所示.于是將I化為先對y后對x的二次積分，得將I化為先對x后對y的二次積分，得(2)積分區(qū)域D如圖9－10 所示. 圖9－9將I化為先對y后對x的二次積分，得I??dx?04xf(x,y)dy I??dy?04y12y4f(x,y)dx.將I化為先對x后對y的二次積分，得圖9－10：I??dx?rr0f(x,y)dy I??dy01rf(x,y)dx(1) ?dy0e1ylnxf(x,y)dx31(2) ?dy?f(x,y)dx001y (3) ?dx?(5) ?dx?0100f(x,y)dy(4) ?dy01f(x,y)dxx2f(x,y)dy??dx?解 (1)因為原積分區(qū)域為Y型區(qū)域,其圖形如圖9－11 所示. 交換積分次序區(qū)域D應視為X型區(qū)域. 故D?(x,y)0?y?1,y?x?1(3?x)20f(x,y)dy??10dyyf(x,y)dx??dx?2f(x,y)(2) 因為原積分區(qū)域D?(x,y)0?y?1,0?x?y為Y型區(qū)域, 其圖形如圖9－12 所示. 交換積分次序區(qū)域D應視為X型區(qū)域. 故 ???01dy?f(x,y)dx??dx?f(x,y)??為X型區(qū)域, 其 (3)因為原積分區(qū)域圖形如圖9－13 所示. 交換積分次序區(qū)域D應視為Y型區(qū)域.圖9－11 圖9－12 D?(x,y)1?x?e,0?y?lnx故 ?1edx?lnx0f(x,y)dy??dy?xf(x,y)（4）因為原積分區(qū)域為Y 型區(qū)域, 其圖形如圖9－14 所示. 交換積分次序區(qū)域D應視為X型區(qū)域. D?(x,y)0?y?1,x? 5dy?0故1f(x,y)dx??dx?110f(x,y)dy.圖9－13 圖9－14（5）因為原積分區(qū)域D?D1?D2,其中D1?(x,y)0?x?1,0?y?x2??1??D2??(x,y)1?x?3,0?y?3－x）?2??為X型區(qū)域, 其圖形如圖9－15 所示. 交換積分次序區(qū)域D應視為Y型區(qū)域.圖9－15 圖9－16故 ?dx?01x20f(x,y)dy??dx?113?2y031(3?x)20f(x,y)dy ??dy6．求由平面x?0、y?0、x?y?1所圍成的柱體被平面z?0與2x + 3y + z = 6所截f(x,y).解 該曲頂柱體如圖9－16所示.V????6?2x?3y?dxdyD7 ??dx??6?2x?3y?dy?.00211習題 9－3，計算下列二重積分：(1) ??(x?y)2sin2?x?y?dxdyD(2?,?)、(?,2?)、.D是頂點為(?,0)、(0,?)的四邊形。(2) ??x2y2dxdy,DD由xy?xy?y?x和y?4x所圍成且x?0、y?0。(3) ??eDyx?ydxdy, D由x軸，y軸和直線x?y?1所圍成。6??1a2b2.解 (1) 積分區(qū)域D如圖9－17所示. D(4) ??(x2a2?y2b2)dxdy,D:x2y21?x??u?v???2??x?y?u?y?1?v?u??2?令?x?y?v，解得?于是原積分區(qū)域D的邊界x?y??、x?y??、x?y???與 圖9－17 x?y?3?、u??、u???相對應. 其積分區(qū)域D’的圖形如圖9－v??、新積分區(qū)域D’的邊界v?3?、18所示. ?x?(x,y)?uJ???(u,v)?y?u因為故 D?x1?v2??y1?2?v121?122 22x?ysin???x?y?dxdy??1???u2sin2v?dudv2D’3?1?2udusin2vdv???2??1?u3???vsin2v?3??????????2?4???2?3???? 圖9－18 ?31?4??(???)?.3223 (2) 積分區(qū)域D如圖9－19所示. ?3??xy?ux?????y?v?y??x令 ?，解得?則新積分區(qū)域D’由u = 1,u = 2,v = 1,v = 4圍成.其積分區(qū)域D’的圖形如圖9－20所示. 圖9－19?x?x?(x,y)?u?vJ???(u,v)?y?y?u?v 因為?u?2)1v?2v7故 ??x2y2dxdy???Du1?uv?dudvv2vD’圖9－20 ?121ududv??2D’v?（3）積分區(qū)域D如圖9－21所示. 411227ududv??1v3ln2. 2?1?x?y?u?x?u?v??y?v令?，解得?y?v則新積分區(qū)域D’由u = v、v = 0和u = 1圍成. 圖9－21 其積分區(qū)域D’的圖形如圖9－22所示.?x?(x,y)?uJ???y?(u,v)?u因為 ?x1?v??y0?v?11?1圖9－22 故 D??eydxdy???e?1?udvd??u?deD’10v1uv0ovd ??u?e?1?du?e?12.（4）積分區(qū)域D如圖9－23所示.?x?arcos??令 ?y?brsin?則新積分區(qū)域為 D’????,r?0???2?,0?r?1? 圖9－23?x?(x,y)?rJ???(r,?)?y?r因為 ?x???y???acos?bsin??arsin??abrbrcos?故 ??(Dxa22?yb22)dxdy???r2abrdrd?D’2? ?ab?1ab?.002 ，求下列區(qū)域D的面積： d??r3dr?133(1)D由曲線xy?xy?xy?5和xy?15所圍成且x?0、y?0.3333(2)D由曲線y?x、y?4x、x?y、x?4y所圍成且x?0、y?0.?u?xy?x?3解 (1) 令?v?xy，解得y?則新積分區(qū)域D’由 u = u = v = v = 15圍成.8其積分區(qū)域D’的圖形如圖9－24所示.?x?x?(x,y)?u?vJ???(u,v)?y?y?u?v 因為??12v 故 SD???dxdyD???815111dudv??du?dv?4?lnv5?’圖9－24?u?????v??(2) 令??yx???x3?x?y?y3，解得? ?則新積分區(qū)域D’由u = u = v = 1和v = 4圍成.其積分區(qū)域D’的圖形如圖9－25所示.?x?(x,y)?uJ???(u,v)?y?u因為 ?x?v?y?v 圖9－25?3???u8v881?8?8?uv8931111???u8v881139?1?(uv)2833?8?8?vu81故 SD???dxdyD33?4411?????uv?2dudv??du?(uv)2dv1188D’1?.188?y?x?0與y?0所圍成，求證：x?y1cos()dxdy?sin1.??x?y2D 證 積分區(qū)域D如圖9－26所示.1?x??v?u???2??x?y?v?y?1?v?u??2?令?x?y?u，解得?則新積分區(qū)域D’由v = 1,v = u, 及v = u圍成. 圖9－26 ??u?4132du 9其積分區(qū)域D’的圖形如圖9－27所示.?x?(x,y)?uJ???(u,v)?y?u因為?x1?v2??y1?2?v121?122故 ??cosDx?yu1dxdy???cos?dudvx?yv2D’ 圖9－271ucosdu0?v2v1vuv??d02v?v??dv?1v??vsin1vd?0112sin1.1?14．選取適當變換，求證：D證 積分區(qū)域D如圖9－28所示. ??f?x?y?dxdy??f?u?du , D:x?y?1.1?x??u?v???2??x?y?u?y?1?u?v??2?令?x?y?v, 解得?則新積分區(qū)域D’由u = 1, u = 1,v = 1及v = 1所圍成其積分區(qū)域D’的圖形如圖9－29所示. 圖9－28?x?x11?(x,y)?u?v122J?????11?(u,v)?y?y2?22?u?v因為故 ??D1f(x?y)dxyd???fu(?udvd2D’1111??du?f(u)dv??f(u)