求下列曲线所围的图形面积

yx=1yx=x=2yx24=+()1yx241=−()yx=yxx=+sin2x=0x=πyx=eyx=−ex=1yx=|ln|y=0x=01.x=10xttyttt=−=−≤≤⎧⎨⎩2202223,,xatyatt==⎧⎨⎩≤≤cos,sin,3302π(8)ra=θ,θθπ==02,(9)ra=e,θθθπ==02,(10)rab=+cosθba≥0(11)rr=3cosθ=+1cosθ(33πθπ≤≤−)(12)θ2cos22ar=(13)ra=cos2θ(14)Descartesxyax333y+=(15)xyaxy44222+=+().12ln23)ln21()1(12221−=−=−=∫xxdxxxA2316)22(2)14()41(22022022=−=⎟⎟⎠⎞⎜⎜⎝⎛−−−=∫∫dyydyyyA32)2cos1(21sin002πππ=−==∫∫dxxxdxA231421)(10−+=−=∫−eedxeeAxx51.0111011.0101101.0)1(ln)1(lnlnlnln−−−=−==∫∫∫xxxxxdxxdxdxxA108110ln1099−=6158)32(2)22)(2(204322032=+−=−−=∫∫dttttdttttA7∫∫−=⋅=206422023)cos(cos12cossin3cos4ππdtttatdttataA2283961516312aaπππ=⎟⎠⎞⎜⎝⎛−=82320223421adaAπθθπ==∫9==∫πθθ202221deaA41()241ae−π10∫∫++=+=ππθθθθθ20222202)cos2cos(21)cos(21dbabadbaA=++=∫πθθπ220222cos12bda2221baππ+11∫∫−−−+=+−=333322)cos22cos43(21])cos1()cos3[(21ππππθθθθθθddAπ=1224022cos214adaA=⋅=∫πθθ132402402221)4cos1(22cos218adadaAπθθθθππ=+=⋅=∫∫14txy=313tatx+=3213taty+=+∞→0:t=A∫∞+′⎟⎠⎞⎜⎝⎛++03321313dttattat=∫∞++−033232)1()21(9dtttta3tu=232=A2032203223)1(3)1(23)1()21(3aduuuaduuua=⎟⎟⎠⎞⎜⎜⎝⎛+−+=+−∫∫∞+∞+θθsin,cosryrx==xyax333y+=θθθθ33cossincossin3+=ar]2,0[πθ∈∫∫+=+=20232220233222tan)1(tantan29)cos(sincossin29ππθθθθθθθθdadaA20232231tan123aa=+⋅−=πθ15θθsin,cosryrx==xyaxy44222+=+()θθ4422cossin+=ar∫∫++=+⋅=2042220442tan1tan1tan2cossin214ππθθθθθθdadaAθtan=t∫∫∞+−−∞++−−=++=0211204222)()(2112ttttdadtttaA201222arctan2attaπ=−=∞+−ya24=x,)0,(axθθsin,cosryarx=+=ya24x=θcos12−=arα∫+−=πααθθαdaA22)cos1(421)(233ααααα42222sincos8)cos1(1)cos1(12)('aaA−=⎟⎟⎠⎞⎜⎜⎝⎛−−+=0)('=αA2πα=2πα0)('αA2πα0)('αA)(αA2πα=)2(πA)2(πA2232222322223102cot)2cot1()cos1(12adada=+−=−=∫∫ππππθθθθ04yx=32/≤≤xxyy=−242ln1≤≤yeyx=lncos02≤≤xaπxatyatt==⎧⎨⎩≤≤cos,sin,3302πxatttyatttt=+=−⎧⎨⎩≤≤(cossin),(sincos),02πra=−(cos1θ)02≤≤θπra=θ,02≤≤θπ3sin3θar=πθ30≤≤1=+=∫40491dxxL2781080−2212111111()()42eeeLyydyyydy−−14+=+−=+=∫∫32001tansecln(tansec)aaLxdxxdxa=+==+∫∫a42043sincos6Lattdxπ==∫a5()cos,()sinxtattytat′′==t2342202Latdtπaπ==∫62222002sin82Lrrdadaππθθθ′=+==∫∫722222001Lrrdadππθθθ′=+=+=∫∫()22412ln241ππππ++++aa833222003sin32Lrrdadaππθθθπ′=+==∫∫13α)2cos1(4sin)cos1(022221αα−=+−=∫adttataL)2cos1(4sin)cos1(222222απα+=+−=∫adttataL123LL=212cos=απα32=)23,)2332((aa−πabABAaBb,hxaybzc2222221++=xya22+=2xza222+=xyza222++=2xxya22+=10()()(226hAaBbhVaxbxdxABabAbahh)Bππ−−=++=+++∫2abcdzczabVccππ34)1(22=−=∫−3yz03022316)(8adxxaVa=−=∫4yz0235∫−−−−−=222222)(xaxxaxdyyxaxAxaxxaaxaxax+−+−−=arcsin)(222222)31(arcsinarcsin)(320022xxadxaxdxxaxxaaa−+=+−∫∫∫+−−+−=aadxxaxaxxaxaxxxa032032)(2)31(arcsin)31(3)45323(a−=π30022154)(adxxaxadxaxaxaxaa=−=−−∫∫3)9832(aV−=πfx()≥0)(0,0xfybxa≤≤≤≤≤y∫=badxxxfV)(2π00≤≤≤≤≤≤αθβπθ,rr()∫=βαθθθπdrVsin)(3231],[babxxxxaPn==210:,{)(0,),(1xfyxxxyxii≤≤≤≤−}y)()(212iiiixfxxV−−≈∆πiiixxfx∆≈)(2π)(max1inix∆=≤≤λ0→λ∫=badxxxfV)(2π236(2)θθcos)(rx=,θθsin)(ry=,ααcos)(ra=,ββcos)(rb=,∫=abdxyV2πααπ22sin)(31ar−ββπ22sin)(31br+∫=abdxy2π∫+baxyd)(312π∫−=αβθθθθπdrrr)sincos'(sin22()∫−++βαθθθθθθπdrrrr332322sincossin2cossin'331∫=βαθθθπdrsin)(323ar≤≤≤≤≤0,0πβθ3cos2222)cos1(32)(cossin3adxxaaaVaaβππββπβ−=−+=∫],[βαβθθθθα==n210)(0,1θθθθrrii≤≤≤≤−33122()(1cos)()(1cos)33iiiiiVrrππθθθ−∆≈−−−θiiirθθθπ∆⋅≈sin)(323∑=∆≈niiiirV13sin)(32θθθπ0)(max1→∆=≤≤iniθλVr=∫233πθθαβ()sindθ1xayb22221+=x2yx=siny=00≤≤xπxy2373xatyatt==⎧⎨⎩≤≤cos,sin,330πxxattyatt=−=−⎧⎨⎩∈(sin),(cos),[,]102πy=0yya=25xyba222+−=()0≤abx6ra=−(cos1θ)7ra=e,θ0≤≤θπ8()(xyaxy222222+=−)x12222234)(abdxxaabVaaππ=−=∫−220221sinπππ==∫xdxV202sin2πππ==∫xdxxV3∫∫==−πππ02732cossin3tdttadxyVaa32097310532)sin(sin6adtttaπππ=−=∫43220236)cos1)(sin(2)(2adttttadxxxfVbaππππ=−−==∫∫3220237)cos1()]cos1(2[)2(adttataaaVππππ=−−−−=∫5()badxxabdxxabxabVaaaa222222222224)()(πππ=−=−−−−+=∫∫−−6303338sin)cos1(32adaVπθθθππ=−=∫733033)1(15sin32aedeaV+==∫ππθπθθπ8∫=40233sin)2(cos34πθθθπdaVθcos=t238∫−=1212323)12(34dttaVπ∫−121232)12(dtt∫−−−=12121221232)12(6)12(21dttttt31−=∫−121232)12(dtt∫−−121212)12(3dtt∫−121232)12(dtt41=∫−−121212)12(43dtt81)21ln(1623122ln1222834121122−+=⎟⎠⎞⎜⎝⎛−+−−−=tttt∫−=1212323)12(34dttaVπ332)12ln(24a⎥⎦⎤⎢⎣⎡−+πyxxa=−()],0[ax∈xac∈[,]xca∫∫−=−caadxaxxdxaxx22022)()(ππ061510254325=−+−caccaaV()ξyxx=+12x∈[,]0ξxaVaV()lim()=→+∞12ξξ239)1(2)1()(22022aadxxxaVa+=+=∫ππ2)(limπξξ=+∞→V21122=+aa1=axayb22221+=xxrrbrxayb22221+=x202221342abdxxaabVaππ=⎟⎠⎞⎜⎝⎛−=∫⎟⎟⎠⎞⎜⎜⎝⎛−−=−+−⋅=∫−232222222222)(34)1(2222rbbaabdxaxbrbbarVarbbaπππ212VV=2213−=brxay=10a2xy=1S1=x2S1a21SS+2x1312131)()(3120221+−=−+−=+∫∫aadxaxxdxxaxSSaa312131)(3+−=aaafaafaaf2)(,21)(2=′′−=′0)(=′af21=a02)21(=′′f21SS+21=a{}12111min()(1)322SSf+==−2πππ)5131154(])([])[(25124042+−=−+−=∫∫aadxaxxdxxaxVaa21=a240V2130π+=12.)(xf]1,0[)1,0(223)()(xaxfxfx+=′a)(xfy=1=x0=y2S1)(xf2aSx1223)()(xaxfxfx+=′23)(axxf=′⎟⎠⎞⎜⎝⎛cxaxxf+=23)(cxxaxf+=223)(223)()(xaxfxfx+=′x2)()(1010adxxfxxdf+=∫∫24)1(af+=ac−=4xaxaxf)4(23)(2−+=2=V1220()(10160)30fxdxaaππ=++∫0=′V5−=a015=′′πV5−=apxy22=0≤≤xaxyx=sin0≤≤xπxxayb22221+=xxatyatt==⎧⎨⎩≤≤cos,sin,330πxra=−(cos1)θθ2cos22ar=(i)(ii)θπ=22411∫∫+=+=aadxpxpdxxppxA00222122ππ=⎥⎥⎦⎤⎢⎢⎣⎡−+2323)2(32ppapπ2∫+=ππ02cos1sin2dxxxA()022cos1ln(coscos1cosππxxxx++++−==)12ln(222++ππ3∫∫−−=−−⋅+−=−aaadxxbaaabdxxaxabxaabA02224222222)(4)(12ππ=⎪⎪⎪⎩⎪⎪⎪