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\centerline
{
\bf Dr. Z's Math251 Handout \#15.3 [Double Integrals over General Regions]
}

By Doron Zeilberger

{\bf Problem Type 15.3a}: 
Evaluate the double integral
$$
\int\int_{D} F(x,y) \, dA \quad ; \quad D=\{ \, (x,y) \, \vert \,
a \leq x \leq b, f(x) \leq y \leq g(x) \,\} \quad.
$$
Or
$$
\int\int_{D} F(x,y) \, dA \quad ; \quad  D=\{ \, (x,y) \, \vert \,
a \leq y \leq b, f(y) \leq x \leq g(y) \,\} \quad.
$$
{\bf Example Problem 15.3a}: 
Evaluate the double integral
$$
\int\int_{D} e^{y^2} \, dA \quad ; \quad D=\{ \, (x,y) \, \vert \,
0 \leq y \leq 1, 0 \leq x \leq y \,\} \quad.
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Set up an {\bf iterated integral} with the
variable ($x$ or $y$) that is the `boss' the {\it outside}
integral and the the other variable the {\it inner} integral.

For the first kind (Type I region) we have
$$
\int\int_{D} F(x,y) \, dA=
\int_{a}^{b} \int_{f(x)}^{g(x)} F(x,y) \, dy \, dx
$$
where the $x$-integral is on the {\it outside} and the
$y$-integral on the {\it inside}.

For the second kind (Type II region) we have
$$
\int\int_{D} F(x,y) \, dA=
\int_{a}^{b} \int_{f(y)}^{g(y)} F(x,y) \, dx \, dy
$$
where the $y$-integral is on the {\it outside} and the
$x$-integral on the {\it inside}.
&
{\bf 1.} 
Here we have a type II region, and the iterated integral is
$$
\int\int_{D} e^{y^2} \, dA=
\int_{0}^{1} \int_{0}^{y} e^{y^2} \, dx \, dy \quad .
$$

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{\bf 2.} Evaluate the iterated integral, by first doing the
inner integral, getting an expression in the outer variable,
and then doing the outer integral.

&
{\bf 2.}  The inner integral is
$$
\int_{0}^{y} e^{y^2} \, dx= e^{y^2} \int_{0}^{y}  \, dx \,
$$
$$
= e^{y^2} \cdot [y-0]= ye^{y^2} \quad .
$$
The outer integral is
$$
\int_{0}^{1} \int_{0}^{y} e^{y^2} \, dx \, dy=
\int_{0}^{1} \left [ \int_{0}^{y} e^{y^2} \, dx \right ]\, dy
$$
$$
=\int_{0}^{1} ye^{y^2} dy= (1/2) e^{y^2} \Bigl \vert_{0}^{1}=
(1/2) [e^{1^2}-e^{0^2}]= (e-1)/2 \quad.
$$
{\bf Ans.:} $(e-1)/2$.
\cr}

{\bf Problem Type 15.3b}: 
Find the volume of the solid
that lies under the surface $z=F(x,y)$ and above the region
bounded by $y=f(x)$ and $y=g(x)$.


{\bf Example Problem 15.3b}: 
Find the volume of the solid
that lies under the plane $x+2y-z=0$ and above the region
bounded by $y=x$ and $y=x^2$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Here we have an extra step of finding the region,
and expressing it either as type I or type II.
It helps to sketch the region.
If the bounding curves are of the form
$y=f(x), y=g(x)$ then it is going to be a type I region.
If the bounding curves are of the form
$x=f(y), x=g(y)$ then it is going to be a type II region.

To get the $a$ and $b$ in $a \leq x \leq b$, we solve
$f(x)=g(x)$, and usually get two roots. These are our
$a$ and $b$. Then by looking at the sketch (or by plugging-in
a random value) decide who is the top and who is at the bottom.
&
{\bf 1.} 
Here the bounding curves of our region are $y=x$ and $y=x^2$.
Setting them equal, we have to solve $x=x^2$ which is the same
as $x-x^2=0$ which is the same as $x(1-x)=0$ and we get
$x=0$ and $x=1$. Now from a diagram (or plug in $x=1/2$ and
see that $(1/2)^2<1/2$) we see that $y=x$ is at the {\bf top}
and $y=x^2$ is at the {\bf bottom}).
So our region  is
$$
D=\{ \, 
(x,y) \, \vert \, 0 \leq x \leq 1 \, , \, x^2 \leq y \leq x \, \} \quad.
$$
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{\bf 2.} Compute the area integral
$$
\int\int_{D} f(x,y) \, dA \quad ,
$$
by first converting it to an iterated intergral, and then
evaluating it step-by-step. First the inner integral, and then
the outer integral.

{\bf Note:}  The answers should {\bf always} be positive.
Volumes are never negative! If you get a negative number it
means that you messed up somewhere.
&
{\bf 2.} Solving for $z$ the plane 
$x+2y-z=0$ is really $z=x+2y$ so our {\bf integrand} is $f(x,y)=x+2y$.
The required volume is
$$
\int\int_{D} (x+2y) \, dA \quad ,
$$
which equals the iterated integral
$$
\int_{0}^{1} \int_{x^2}^{x} (x+2y) \, dy \,dx \quad .
$$
First we do the inner integral
$$
 \int_{x^2}^{x} (x+2y) \, dy= xy+y^2 \Bigl \vert_{x^2}^{x}
$$
$$
=x\cdot x+x^2  - (x\cdot x^2+(x^2)^2)=
2x^2-x^3-x^4 \quad.
$$
Having done the inner integral, we are ready for the outer intgeral
$$
\int_{0}^{1} \int_{x^2}^{x} (x+2y) \, dy \,dx 
$$
$$
=\int_{0}^{1} \left [ \int_{x^2}^{x} (x+2y) \, dy \, \right ] \,dx 
$$
$$
=\int_{0}^{1} (2x^2-x^3-x^4) \,dx 
$$
$$
=2x^3/3-x^4/4-x^5/5 \Bigl \vert_{0}^{1}=
2/3-1/4-1/5=13/60 \quad .
$$
{\bf Ans.:} $13/60$.
\cr}

{\bf Problem Type 15.3c}: Sketch the region of integration
and change the order of integration.
$$
\int_{a}^{b} \int_{f_1(x)}^{f_2(x)} F(x,y) \, dy \, dx
$$
where $F(x,y)$ is not given specifically (i.e. is left as
an abstract function).

{\bf Example Problem 15.3c}: Sketch the region of integration
and change the order of integration.
$$
\int_{0}^{1} \int_{4x}^{4} F(x,y) \, dy \, dx
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Sketch the type I region
$$
D=\{ \, (x,y) \, \vert \, a \leq x \leq b \, , \, 
f_1(x) \leq y \leq f_2(x) \, \}
$$
and express it as a type II region
$$
D=\{ \, (x,y) \, \vert \, c \leq y \leq d \, , \, 
g_1(y) \leq x \leq g_2(y) \, \}
$$

&
{\bf 1.} The region of integration, $D$ is
$$
D=\{ \, (x,y) \, \vert \, 0 \leq x \leq 1 \, , \, 
4x \leq y \leq 4 \, \} \quad .
$$
This is a triangle whose vertices are $(0,0)$
$(0,1)$ and $(1,4)$. Viewing it from the point of
view of the $y$-axis, the {\bf main road} is
$0 \leq y \leq 4$ and the ``side streets'' are
horizontal line-segments stretching from $x=0$ to $x=y/4$.
So our region $D$ written as a type II integral is:
$$
D=\{ \, (x,y) \, \vert \, 0 \leq y \leq 4 \, , \, 
0 \leq x \leq y/4 \, \} \quad .
$$
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{\bf 2.} Set-up the iterated integral
$$
\int_{c}^{d} \int_{g_1(x)}^{g_2(x)} F(x,y) \, dx \, dy \quad .
$$
Of course you can't evaluate it, since you were not
given $F(x,y)$ .

&
{\bf 2.} 
$$
\int_{0}^{4} \int_{0}^{y/4} F(x,y) \, dx \, dy \quad .
$$
This is the {\bf Ans.} .

\cr}

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