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\hfill \thepage} %} \input{tcilatex} \begin{document} \subsubsection{MA 116\qquad \qquad \qquad \qquad \qquad EXAM 4\qquad \qquad \qquad \qquad \qquad MISC.\#1} \vspace{1pt} \subsubsection{Solutions} \vspace{1pt} 1. (a) Evaluate $\int_{0}^{1}\int_{x^{2}}^{x}\left( 3x-y\right) dydx$ \ \ \ \ (b) Reverse the order of integration of the iterated integral in part (a). \vspace{1pt} Solution: (a) $\int_{0}^{1}\int_{x^{2}}^{x}\left( 3x-y\right) dydx=\int_{0}^{1}\left[ \int_{x^{2}}^{x}\left( 3x-y\right) dy\right] dx=\int_{0}^{1}\left[ 3xy-% \tfrac{1}{2}y^{2}\right] _{y=x^{2}}^{y=x}dx$ $=\int_{0}^{1}\left[ \left( 3x^{2}-\tfrac{1}{2}x^{2}\right) -\left( 3x^{3}-% \tfrac{1}{2}x^{4}\right) \right] dx=\int_{0}^{1}\left[ \left( \tfrac{5}{2}% x^{2}\right) -\left( 3x^{3}-\tfrac{1}{2}x^{4}\right) \right] dx$ $=\left( \tfrac{5}{6}x^{2}-\tfrac{3}{4}x^{4}+\tfrac{1}{10}x^{5}\right) _{0}^{1}=\tfrac{5}{6}-\tfrac{3}{4}+\tfrac{1}{10}=\frac{11}{60}$ (b) $\int_{0}^{1}\int_{y}^{\sqrt{y}}\left( 3x-y\right) dxdy$ (see graph below) \qquad \qquad \qquad \qquad \qquad \FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{% \special{language "Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0in;display "USEDEF";function \TEXUX{$x$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$x^{2}$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax "5";xviewmin "0";xviewmax "3";yviewmin "0";yviewmax "3";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}} \vspace{1pt} 2. Find the volume under $x+y-z=0$ and over the region bounded by the x-axis, the line $x=1$ and the parabola $y=x^{2}.$ \vspace{1pt} Solution: \qquad \qquad \qquad \qquad \qquad \FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{% \special{language "Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0in;display "USEDEF";function \TEXUX{$x^{2}$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$\left( 1,0\right) ,\left( 1,1\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax "3";xviewmin "0";xviewmax "3";yviewmin "0";yviewmax "3";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}} $V=\int_{0}^{1}\int_{0}^{x^{2}}\left( x+y\right) dydx=\int_{0}^{1}\left[ \int_{0}^{x^{2}}\left( x+y\right) dy\right] dx=\int_{0}^{1}\left[ \left( xy+% \tfrac{1}{2}y^{2}\right) \right] _{0}^{x^{2}}dx$ $=\int_{0}^{1}\left( x^{3}+\tfrac{1}{2}x^{4}\right) dx=\left( \tfrac{1}{4}% x^{4}+\tfrac{1}{10}x^{5}\right) _{0}^{1}=\left( \tfrac{1}{4}+\tfrac{1}{10}% \right) =\tfrac{7}{20}$ \vspace{1pt} 3. Consider $\int_{1}^{2}\int_{1}^{x}\tfrac{x^{2}}{y^{2}}dydx$ \ \ (a) Write the equivalent iterated integral in the reverse order. \ \ (b) Evaluate the integral obtained in (a). \vspace{1pt} Solution: (a) $\int_{1}^{2}\int_{y}^{2}\tfrac{x^{2}}{y^{2}}dxdy$ (see graph below) \qquad \qquad \qquad \qquad \qquad \FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{% \special{language "Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0in;display "USEDEF";function \TEXUX{$\left( 1,1\right) ,\left( 2,2\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$\left( 1,1\right) ,\left( 2,1\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$\left( 2,1\right) ,\left( 2,2\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax "2";xviewmin "0";xviewmax "3";yviewmin "0";yviewmax "3";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}} (b) $\int_{1}^{2}\int_{y}^{2}\tfrac{x^{2}}{y^{2}}dxdy=\int_{1}^{2}\left[ \int_{y}^{2}\tfrac{x^{2}}{y^{2}}dx\right] dy=\int_{1}^{2}\left[ \tfrac{1}{% 3y^{2}}x^{3}\right] _{x=y}^{x=2}dy$ $=\int_{1}^{2}\left[ \tfrac{8}{3y^{2}}-\tfrac{1}{3}y\right] dy=\left[ \tfrac{% -8}{3y}-\tfrac{1}{6}y^{2}\right] _{1}^{2}=\left[ \left( \tfrac{-8}{6}-\tfrac{% 4}{6}\right) -\left( \tfrac{-8}{3}-\tfrac{1}{6}\right) \right] $ $=\left[ \tfrac{-12}{6}+\tfrac{17}{6}\right] =\tfrac{5}{6}$ \vspace{1pt} 4. Give two expressions for the $\int \int_{D}xy\;dA$ where \textit{D} is the region in the figure below. Do \textit{not }evaluate the expressions. \vspace{1pt}\qquad \qquad \FRAME{dtbpF}{2.2658in}{2.1741in}{0pt}{}{}{% img00002.gif}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.2658in;height 2.1741in;depth 0pt;original-width 164.875pt;original-height 158.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/AH87CVJL/img00002.gif';file-properties "XNPEU";}} \vspace{1pt} Solution: Solving for the point of intersection we get: \vspace{1pt}$\sqrt{x}=x^{2}\rightarrow x=1\rightarrow y=1$ therefore the point of intersection is $\left( 1,1\right) $ Integrating with respect to $y$ and then with respect to $x$ we get:\ $\dint_{0}^{1}\dint_{x^{2}}^{\sqrt{x}}xy\;dydx$ Since $y=\sqrt{x}$, this implies that $x=y^{2}$. \ Similarly, \ $y=x^{2}$, implies that $x=\sqrt{y}$. \ We already found the point of intersection, namely $(1,1)$. \ Integrating with respect to $x$ and then with respect to $y$ we get: $\dint_{0}^{1}\dint_{y^{2}}^{\sqrt{y}}xy\;dxdy$ \vspace{1pt} 5. Consider $r=f\left( \theta \right) =\cos 2\theta $ (a) Sketch $f$ (b) Find the total area enclosed by $f.$ (c) Set up an integral to evaluate the total length of $f$ (DO NOT INTEGRATE) \vspace{1pt} Solution: (a) \qquad \qquad \qquad \qquad \qquad \FRAME{itbpF}{3in}{2.0003in}{0in}{}{}{}{% \special{language "Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0in;display "USEDEF";function \TEXUX{$\cos 2\theta $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "-3.1416";xmax "3.1416";xviewmin "-1.040000";xviewmax "1.040775";yviewmin "-1.039976";yviewmax "1.040799";rangeset"X";phi 45;theta 45;plottype 8;numpoints 49;axesstyle "normal";xis \TEXUX{v952};var1name \TEXUX{$\theta $};}} (b) $\cos 2\theta =0\rightarrow \theta =\tfrac{\pi }{4}$ therefore we can calculate the area of the top part of the region that lies along the x-axis in the first quadrant and multiply by $8.$\newline $A=8\cdot \tfrac{1}{2}\int_{0}^{\tfrac{\pi }{4}}\left( \cos 2\theta \right) ^{2}d\theta =4\int_{0}^{\tfrac{\pi }{4}}\left( \cos ^{2}2\theta \right) d\theta =4\int_{0}^{\tfrac{\pi }{4}}\tfrac{1}{2}\left( 1+\cos 4\theta \right) d\theta $ $=2\int_{0}^{\tfrac{\pi }{4}}\left( 1+\cos 4\theta \right) d\theta =2\left[ \theta +\tfrac{1}{4}\sin 4\theta \right] _{0}^{\tfrac{\pi }{4}}=2\left[ \left( \tfrac{\pi }{4}\right) \right] =\tfrac{\pi }{2}$ (c) $L=8\int_{0}^{\tfrac{\pi }{4}}\sqrt{\cos ^{2}2\theta +4\sin ^{2}2\theta }% d\theta $ \vspace{1pt} 6. Let $S$ be the region bounded by the surface $z=xy,$ the cylinders $% y=x^{2}$ and $y^{2}=x,$ and the plane $z=0.$ Find the volume of $S.$ \vspace{1pt} Solution: We are considering the region bounded above by $z=xy$ below by the xy-plane and by the cylinders $y=x^{2},y^{2}=x.$ The region in the xy-plane look like: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \FRAME{itbpF}{3in}{% 2.0003in}{0in}{}{}{}{\special{language "Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0in;display "USEDEF";function \TEXUX{$x^{2}$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$\sqrt{x}$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax "2";xviewmin "0";xviewmax "2";yviewmin "0";yviewmax "2";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}} $V=\int_{0}^{1}\int_{x^{2}}^{\sqrt{x}}xydydx=\int_{0}^{1}\left[ \tfrac{xy^{2}% }{2}\right] _{x^{2}}^{\sqrt{x}}dx=\tfrac{1}{2}\int_{0}^{1}\left( x^{2}-x^{5}\right) dx=\tfrac{1}{12}$ \end{document} %%%%%%%%%%%%%%%%% End /document/ExamIV_Misc#1_Sol.tex %%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% Start /document/AH87CVJL/img00002.gif %%%%%%%%%%%%%%%% GedQxdSX[C`t@\O@@@@@@@H@@@@`@@H`@@@@@BH@@B@`@BLp@CLw@[jrpC@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@|~pCJhdBH`@~O@@@p@|@@@@O@CpGR~D@@@@\ O@l@@@@@pv@HM@@``@|_B\`Dl`F|`HLaJ\aLlaN|aPLbR\bTlbV|bXLcZ\c\lc^|c`Ldb\ddld f|dhLej\ellen|epLfr\ftlfv|fxLgz\g|lg~|g@MhB]hDmhF}hfD@@X`JUaMU`OEiR}eOUiQm` V]iLUkSmkj|jWu``ek^mldLlaEkd}_iulnech]mj]nomn]LnN]mmun|]bxMamoBfbA~`E^pHva Gfn[MqNVqzelceqOnrH\olUnT~nV^hXna_vbQUs]FgaFpdFcKekintnFbl~urnbZujsnfgFfq~v sT_zkg{fwntwzC_yKhmVxctwx[@Mx@MyJc~vyVGzQ_zhlxVoyFWhRWkgQfw]gc}lK^js]oW~ sc~toqjs~y[^bF^xwPrSy{~wc_JJ_ew~u_~wEQkg@f`DxO]_FHAi_IH}m`LhC~PAx~i`HUCb_ DGFzauwQmabGD~aChTIbSg~DakhD^bUhA]alxJNazfdeZhxma`eHKRcTeMBXoh@QbrhGvcuHTuZ xhkib{XSN@FiObdNUEFcNIFBeuXhidOvPZeQHVzcZIGrdDYYz]`yQnP[y`icfYYbfQiZNbmiW~a piKe[ki]QfuiY^gkdmAZJWUzfMy_rSziYQgEz[zVIZWueLzN^hHjlQh}ia~VOjQEiXJQRiUZgqg RJiBj{tfJTWJizdfzib\YeYuAzisJHjiHwEuikj_rWhZRYQcJmVgmZhVRSVDidjZpvh|zIMYpZ RNQyjrNiBVorPUEJV[lFielzjWNl@GJamHYlnUUUcmWKztVh`KqfTAJhU[XUBin^kdvnn[qBoBz OfVu{zfiHfuFoRthmmOx@XokK~foz[AG^LI@kkL\CORyF@Z]JlBSqnz{NpBI_JqXJGwqGU[jqtJ IkS_|vQgbLcjrkLgnbh|jFsyDKSRI{K[ewLNOT[BssK{cesaTPcOCMuZYFmGImYlROceVSsQ@K KFuWikQuZdUau\Oku}TW_u_}g~Z[MnRje]Izsa[Num^HVw]mm[_SHMMKwOghMp~Z_v}]Gwk]FR [Z[HZScmWFxfx`KeFEZ[ah\^wnW|Ow[u\bOw[tdSyqtaKirMxjy[~Kuy}yekwYThKzmDi{yf~pf zj~JqzenRCzN~YMzI^msx|Ncame[Q]{pxe_yPgsBkRIlWwjtlojC}yMnyepKnlTs{iINtSxYFps w^utCwJLOw_@OjM}M[J}wv{ZS`doamZf_qn]ymsIrShV^pfecg~J[Y[pOnURv_ZlZvBj{uYhZB piuBsAutR`CPeE]se@P~{_B\`Dl`F|`@x`J\aLT_x[@OLbP\bRlbT|bVLcPTc\DcPL_LLdb\ddl df|dDLahddNtc]|enLfqDfQDdk|fxLgzDa{TamdfsLhoDh\tf}|hHMiFTeJ}_zShPeh^LjqthM} jXEeLMiOUj]D@`Ml@Hj^}bVMkhMmK\k}T_]ejoDlTTnpDnX|ljMoyemz|gv}k@hbt]femfz}pGf oxtoEveC~a`mnEFoHnrMUqkdqMfQ^nc]herV^tGEsgTsMNcO^@]nc[VtdNvxttf|mifsAvsBm lpVv~~dtfe[fhknurTj}~wJ_dAWdv^xAuxFgcH_yVfMOdhFzjVlSxPMyW~qot\n}[c^JBTqS ~vKplopgg~yOqp{~{GRx_Pys}~gQXr^xGyi_EDQ_tgPE`BxuQ`CdAz_PXD}RJxBZ]MhSEaYhFM STXEn\MxC^`QXQyaex@Q`ahFBaUeIZbhhHBcXdFuamhoi_ggJb`VDMVcrvMfc@iDZV}hKJ\qxQV _UDORddVvebHIzidLYIzdPYPN@ST~ldSyhu\xhUFeCpVre]is]eViXRfEYC]fgyXjfVFm}dmymA gkiOAfgi]ReqW^Beoy^fXqG]Fd@Z`jWXv\zgavazaIzbBcFZcnW{GdjaRzdjVJfbRhDEffiiebu iWJhzhGdeN__ZibV|Egbjndj~{jfUmzkfZqjlbUluSMjwfVbkfJAyPuj]]kZnV`Xypn^FK_z`[ DoRlL[sJPmDtV]RKpRmOkrrbX[v~VVk}ym_kPrm{xxvl~UxvQEKz^UjdmYnTdozn@kzJoxS{Voo kvVnkkXjo{{eQmRt~PAl@OpXh~jH\BkTItsmnt{CsgLVmzpULDopIyhRqZlEWQVZFCr[lD{g_ \ISerW^frj|J[QnIIrWFKCdnLM{bZVNbsylNKcYXO{sz\Lgjs|POkvlmJtH]Kf[b\h]tM}^MU^ JSOuLySk^AeV^uX}T~YWt]uu^}WSfZ]MQvRmYoR^g[vp]\KwsM]Wwv}]cwym^kwFRdWH[Ghbu{ }wc{VeP~}`kfE~\xHnHqxTzInxMnB__J~cOy}hecyL[`Ky[Nay`~YsT`Uh{xpYiWzf^~Ygauj _zhnjzkUkK{sNOexF~e[v{nrI{v~mGyAolv{CBV|FO~a|Io{m|L@~|YjsG}VHuwhSou[X_v {c\^n}^_YF~I~xG[AD@lC %%%%%%%%%%%%%%%%% End /document/AH87CVJL/img00002.gif %%%%%%%%%%%%%%%%%