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\centerline
{
\bf Dr. Z's Math251 Handout \#15.6 (2nd ed.)
[Change of Variables in Multiple Integrals]
}

By Doron Zeilberger

{\bf Problem Type 16.5a}: 
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 16.5a}: 
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 15.6b}: 
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 15.6b}:  
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





