%ho155.tex (2nd ed.) %%a Plain TeX file by Doron Zeilberger for Math 251(1 page) %\bigskip & \bigskip \cr %& %\cr %begin macros \def\L{{\cal L}} \baselineskip=14pt \parskip=10pt \def\Tilde{\char126\relax} \font\eightrm=cmr8 \font\sixrm=cmr6 \font\eighttt=cmtt8 \magnification=\magstephalf \def\P{{\cal P}} \def\Q{{\cal Q}} \parindent=0pt \overfullrule=0in \def\frac#1#2{{#1 \over #2}} %end macros \centerline { \bf Dr. Z's Math251 Handout \#15.5 (2nd ed.) [Change of Variables in Multiple Integrals] } By Doron Zeilberger {\bf Problem Type 15.5a}: Find the Jacobian of the transformation $$ x=g(u,v,w)\quad, \quad y=h(u,v,w) \quad, \quad z=k(u,v,w) . $$ {\bf Example Problem 15.5a}: Find the Jacobian of the transformation $$ x=u^2 v \quad, \quad y=v^2w \quad, \quad z=w^2u . $$ \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\bf Steps} & {\bf Example} \cr & \cr {\bf 1.} Compute all the entries in the Jacobian matrix $$ \left \vert \matrix{ {{\partial x} \over {\partial u}} & {{\partial x} \over {\partial v}} & {{\partial x} \over {\partial w}} \cr {{\partial y} \over {\partial u}} & {{\partial y} \over {\partial v}} & {{\partial y} \over {\partial w}} \cr {{\partial z} \over {\partial u}} & {{\partial z} \over {\partial v}} & {{\partial z} \over {\partial w}} } \right \vert \quad . $$ & {\bf 1.} $$ \left \vert \matrix{ 2uv & u^2 & 0 \cr 0 & 2vw & v^2 \cr w^2 & 0 & 2uw } \right \vert \quad . $$ \cr \bigskip & \bigskip \cr {\bf 2.} Evaluate the determinant: $$ \left \vert \matrix{ {{\partial x} \over {\partial u}} & {{\partial x} \over {\partial v}} & {{\partial x} \over {\partial w}} \cr {{\partial y} \over {\partial u}} & {{\partial y} \over {\partial v}} & {{\partial y} \over {\partial w}} \cr {{\partial z} \over {\partial u}} & {{\partial z} \over {\partial v}} & {{\partial z} \over {\partial w}} } \right \vert \quad $$ $$ = ({{\partial x} \over {\partial u}}) \left \vert \matrix{ {{\partial y} \over {\partial v}} & {{\partial y} \over {\partial w}} \cr {{\partial z} \over {\partial v}} & {{\partial z} \over {\partial w}} } \right \vert \quad - ({{\partial x} \over {\partial v}}) \left \vert \matrix{ {{\partial y} \over {\partial u}} & {{\partial y} \over {\partial w}} \cr {{\partial z} \over {\partial u}} & {{\partial z} \over {\partial w}} } \right \vert \quad $$ $$ + ({{\partial x} \over {\partial w}}) \left \vert \matrix{ {{\partial y} \over {\partial u}} & {{\partial y} \over {\partial v}} \cr {{\partial z} \over {\partial u}} & {{\partial z} \over {\partial v}} } \right \vert \quad $$ & {\bf 2.} $$ = 2uv \left \vert \matrix{ 2vw & v^2 \cr 0 & 2wu } \right \vert \quad - u^2 \left \vert \matrix{ 0 & v^2 \cr w^2 & 2wu } \right \vert \quad $$ $$ + 0 \cdot \left \vert \matrix{ 0 & 2vw \cr w^2 & 0 } \right \vert \quad $$ $$ = 2uv [(2vw)(2uw)-0]-u^2[0-(v^2)(w^2)]+0 $$ $$ =9u^2v^2w^2 \quad. $$ {\bf Ans.}: $9u^2v^2w^2$. \cr} {\bf Problem Type 15.5b}: Use the given transformation to evaluate the integral $$ \int\int_{R} F(x,y) \, dA \quad , $$ where $R$ is the triangular region with vertices $(p_1,p_2)$,$(q_1,q_2)$, $(r_1,r_2)$; $x=au+bv$, $y=cu+dv$. {\bf Example Problem 15.5b}: Use the given transformation to evaluate the integral $$ \int\int_{R} (x+y) \, dA \quad , $$ where $R$ is the triangular region with vertices $(0,0)$,$(2,1)$, $(1,2)$; $x=2u+v$, $y=u+2v$. \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\bf Steps} & {\bf Example} \cr & \cr {\bf 1.} Figure out the region in the $uv$-plane that gets transformed. Since a triangle goes to a triangle, we need to find the 3 vertices. Solve for $u$, $v$ in terms of $x,y$ and find the three points. Call the new triangle $R'$. & {\bf 1.} Since $x=2u+v$, $y=u+2v$, when $(x,y)=(0,0)$ $u=0,v=0$ so the point $(0,0)$ goes to the point $(0,0)$. When $(x,y)=(1,2)$, we have to solve the system $1=2u+v,2=u+2v$ giving us $u=0,v=1$ so $(1,2)$ goes to $(0,1)$. Similarly, $(2,1)$ goes to $(1,0)$. So the region in the $uv$-plane is the far simpler triangle whose vertices are $(0,0),(1,0),(0,1)$. Let's call this region $R'$. \cr \bigskip & \bigskip \cr {\bf 2.} Find the Jacobian of the transformation. In this case of a so-called linear transformation, the Jacobian is simply $ad-bc$. Also express $F(x,y)$ in terms of $(u,v)$ using the transformation. $$ \int\int_{R} F(x,y) \, dA= $$ $$ \int\int_{R'} F(au+bv,cu+dv) |(ad-bc)| \, dA \quad . $$ (Note that one must take the {\bf absolute value} of the Jacobian) & {\bf 2.} The Jacobian is $(2)(2)-(1)(1)=3$, so $$ \int\int_{R} (x+y) \, dA= \int\int_{R'} (2u+v+u+2v) \cdot |3| \, dA= $$ $$ 9 \int\int_{R'} (u+v) \, dA \quad . $$ \cr \bigskip & \bigskip \cr {\bf 3.} Draw the region (in this case triangle) in the $uv$- plane and express it as a type I (or type II) region. Then set-up the appropriate iterated integral, by deciding on the {\bf main road} and the {\bf side streets}. & {\bf 3.} The region is the triangle bounded by the axes and the line $u+v=1$. It can be written as $$ \{ (u,v) \, \vert \, 0 \leq u \leq 1 \, , \, 0 \leq v \leq 1-u \, \} \quad. $$ Our area-integral is thus equal to the iterated integral $$ 9 \int_{0}^{1}\int_{0}^{1-u} (u+v) \, dv \, du \quad . $$ The inner integral is $$ \int_{0}^{1-u} (u+v) \, dv= uv+{{v^2} \over {2}} \Bigl \vert_{0}^{1-u} $$ $$ =u(1-u)+{{(1-u)^2} \over {2}}=(1-u^2)/2 \quad , $$ and the whole integral is $$ {{9} \over {2}} \int_{0}^{1} (1-u^2)\, du= {{9} \over {2}} \left [ u- {{u^3} \over {3}} \right ]\Bigl \vert_{0}^{1}= {{9} \over {2}} \cdot {{2} \over {3}}=3 \quad . $$ {\bf Ans.}: $3$. \cr} {\bf A Problem from a previous Final} Find the Jacobian of the transformation $$ x=u+v+w \quad, \quad y=u^2+v^2+w^2 \quad , \quad z=u^3+v^3+w^3 \quad . $$ Simplify as much as you can! {\bf Ans.}: $6(vw^2 - v^2 w -u w^2+u^2 w+ uv^2 -u^2 v$. {\bf Another Problem from a Previous Final} Use the transformation $$ x=2u+v \quad , \quad y=u+2v \quad , $$ to evaluate the integral $$ \int \, \int_R \, (2x-y) \, dA $$ where $R$ is the triangular region with vertices $(0,0)$, $(2,1)$, and $(1,2)$. {\bf Ans.}: ${{3} \over {2}}$. \end