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\centerline
{
\bf Dr. Z's Math251 LAST Handout  [The Meaning of It All]
}

By Doron Zeilberger

{\bf Concept 1: Indefinite Integral}

{\bf Notation}: $\int f(x) \, dx $ \quad \quad .

{\bf Input}: A {\it function} $f(x)$ of the single-variable $x$.

{\bf Output}: A {\bf function} $F(x)$ \quad \quad .

{\bf Meaning:} $F(x)$ is a function such that  $F'(x)=f(x)$.
Since there are many such functions (for example if $f(x)=x$,
then $F(x)=x^2/2$, $F(x)=x^2/2+1$,  $F(x)=x^2/2+11$, etc.
all have this property, we write $+C$ for {\it arbitrary} constant,
to indicate that this is really a {\it family of answers}.

{\bf How to compute it:} Memorize the answer for basic functions
and use {\bf integration techniques} for complicated ones.
Better still: use Maple (but not on exams!).

{\bf Concept 2: Definite Integral}

{\bf Notation}: $\int_a^b f(x) \, dx $ \quad \quad .

{\bf Input}: A {\it function} $f(x)$ of the single-variable $x$,
and two {\it numbers} $a$ and $b$.

{\bf Output}: A {\bf number}. 

{\bf How to Compute it}: Find the
{\it indefinite} integral $\int f(x) \, dx$, let's call
it $F(x)$, and do $F(b)-F(a)$.

{\bf Meanings}: 1. If $f(x)$ is positive 
for $a \leq x \leq b$, then it is the area under the curve $y=f(x)$,
above the $x$ axis, and between the vertical lines $y=a$ and $y=b$.

2. If you have a one-dimensional wire stretched out from
$x=a$ to $x=b$, and its (linear) density-function is
$f(x)$, then this is its {\bf mass}.

{\bf Important Special case:} when $f(x)=1$, $\int_{a}^{b} dx=b-a$,
the length of the interval $b-a$.

{\bf Concept 3: Line-Integral of the arclength kind}

{\bf Notation}: $\int_C f(x,y,z) ds$ \quad \quad .

{\bf Inputs}: a function of {\bf three} variables $(x,y,z)$
(or sometimes {\bf two} $(x,y)$)) and a {\bf curve}
described (usually) in {\bf parametric} representation.
$x=Expression_1(t) , y=Expression_2(t), z=Expression_3(t)$
($a \leq t \leq b$) or equivalently in {\bf vector notation}
${\bf r}(t)= \langle x(t), y(t), z(t) \rangle$ ($a \leq t \leq b$).

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: replace $ds$ by $|{\bf r}'(t)|dt$,
$\int_C$ by $\int_a^b$ and $f(x,y,z)$ by
$f(x(t),y(t),z(t)$, getting a definite integral in $t$.

{\bf Meanings}: 1. If you bulid your two-dimensional house along the
curve, with the floor being the curve and the ceiling-height
given by the function $f(x,y,z)$, it is the area of that house.

2. If it is a wire whose linear density is given by
$f(x,y,z)$ then it is its mass \quad \quad .

{\bf Important Special case:} If $f(x,y,z)=1$ we get
the {\bf length} of the curve (also called {\bf arclength}).


{\bf Concept 3': Line-Integral of the Vector-Field kind}

{\bf Notation}: $\int_C {\bf F} \cdot  d{\bf r} $

{\bf Inputs}: a {\bf vector-field} of {\bf three} variables $(x,y,z)$,
${\bf F}= \langle P,Q,R \rangle$,
(or sometimes {\bf two} $(x,y)$ and 
${\bf F}=\langle P,Q \rangle$) and a {\bf curve}
described (usually) in {\bf parametric} representation.
$x=Expression_1(t) , y=Expression_2(t), z=Expression_3(t)$
($a \leq t \leq b$) or equivalently in {\bf vector notation}
${\bf r}(t)= \langle x(t), y(t), z(t) \rangle$ ($a \leq t \leq b$).

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: replace $d {\bf r}$ by $ {\bf r}'(t)dt$,
$\int_C$ by $\int_a^b$ and ${\bf F}$ by
what you get by plugging-in in $P,Q,R$ the expressions
in terms of $t$ for $x,y,z$. Then compute
the dot-product ${\bf F} \cdot {\bf r}'(t) dt$ and
integrate from $t=a$ to $t=b$. Equivalently
do $\int P dx + Q dy + R dz $ with
$dx=x'(t)dt$, $dy=y'(t)dt$, $dz=z'(t)dt$.

{\bf Meanings}: Several in physics. Most important,
the {\bf work} done by a {\bf force} ${\bf F}$ moving along $C$.


{\bf Important Special case:} None.


{\bf Concept 4: Iterated Integral of two variables}

{\bf Notation}: 
$\int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx \quad \quad$ (type I)

or

$\int_{a}^{b} \int_{g_1(y)}^{g_2(y)} f(x,y)dx dy \quad \quad $ (type II)

{\bf Inputs}: a {\bf function} $f(x,y)$ of two variables,
{\bf numbers} $a$ and $b$ and {\bf functions} (that sometimes
happen to be constant functions hence may be numbers)
$g_1(x),g_2(x)$ (for type I) or $g_1(y),g_2(y)$ (for type II).

{\bf Output}: a {\bf number}.

{\bf Watch out for nonsense:} $\int_x^{2x} \int_{0}^{1} f(x,y) dy dx$
is utter-nonsense ({\bf outside} limits-of-integration can only
be numbers).

{\bf How to Compute it}: First you do the {\bf inner-integral},
getting a function of the outside variable-of-integration, then
you do the {\bf outside-integral} getting a number.

{\bf Meanings}: these come up when evaluating double-integrals
(alias area-integrals).

{\bf Important Special case:} None.


{\bf Concept 5: Iterated Integral of three variables}

Analogous to two variables, but now you have an
{\bf inside} integral, a {\bf middle} integral, and an
{\bf outside} integral.

{\bf Concept 6: Double Integral (alias Area-integral) } 

{\bf Notation}: 
$\int\int_D f(x,y)\, dA$ \quad \quad .

{\bf Inputs}: a {\bf function} $f(x,y)$ of two variables,
and a {\bf region} $D$ in the plane ($xy$-plane).

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: 
Express the region $D$ either in type-I style, or
type-II style, or polar.
Then convert the double-integral into an {\bf iterated integral}.

{\bf Meanings}: 1. If $f(x,y)$ is positive, and you build
a building whose floor is the region $D$, and whose
ceiling-height above a typical point $(x,y)$ is given
by the function $f(x,y)$, then it is the {\bf volume} of
that building.

2. If you have a plate whose shape is $D$ and $f(x,y)$ is its
(surface) {\bf density function}, then it is the {\bf mass} of the plate.

{\bf Important Special case:} if $f(x,y)$ is $1$ then
you get the {\bf area} of $D$.


{\bf Concept 7: Surface Integral of the scalar kind}

{\bf Notation}: 
$\int\int_S f(x,y,z)\, dS$ 

{\bf Inputs}: a {\bf function} $f(x,y,z)$ of three variables,
and a {\bf surface} $S$ 
(either given {\bf explicitly} as $z=g(x,y)$ or
{\bf parametrically} as as ${\bf r}(u,v)$.

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: 
In the explicit case: find the {\bf projection} on
the $xy$-plane, call it $D$, then replace
$dS$ by $\sqrt{1+(g_x)^2+(g_y)^2} dA$ 
and do the resulting double-integral
(not forgetting to replace $z$ by $g(x,y)$ so you
won't get any mention of $z$.)

In the parametric case: Let $D$ be the
{\bf parameter-space} region (usually given by
the problem). Replace $x,y,z$ by their expression
in $(u,v)$ and replace $dS$ by $|r_u \times  r_v | dA$,
and do the resulting double-integral.

{\bf Meanings}: 1. If you have
a building whose floor is the surface $S$, and whose
ceiling-height above a typical point $(x,y,z)$ is given
by the function $f(x,y,z)$, then it is the {\bf volume} of
that building.

2. If you have a piece of material that fits on S and
$f(x,y,z)$ is the density function, then it is the {\bf mass}.

{\bf Important Special case:} if $f(x,y,z)$ is $1$ then
you get the {\bf surface-area} of $S$.

{\bf Concept 7': Surface Integral of the vector-field kind}

{\bf Notation}: 
$\int\int_S {\bf F}\, \cdot \, d{\bf S}$  \quad \quad .

{\bf Inputs}: a {\bf vector-field} ${\bf F}(x,y,z)$ of three variables,
and three components (called $P,Q,R$)
and a {\bf surface} $S$ 
(either given {\bf explicitly} as $z=g(x,y)$ or
{\bf parametrically} as as ${\bf r}(u,v)$ ) .

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: 
In the explicit case, $z=g(x,y)$ find the projection of $S$ on the
$xy$-plane, call it $D$,
and find the double-integral
$$
\int \int_D 
\left ( -P {{\partial z} \over {\partial x}}
-Q {{\partial z} \over {\partial y}}+R \right ) \, dA
$$

In the parametric case: Let $D$ be the
{\bf parameter-space} region (usually given by
the problem). replace $x,y,z$ by their expression
in $(u,v)$ and replace $d{\bf S}$ by $(r_u \times r_v) dA$,
take the dot-product and do the resulting double-integral.

{\bf Meanings}: Several in physics (most notably
Electricity and Magnetism).


{\bf Important Special case:}  None.


{\bf Concept 8.: Triple Integral (alias Volume-Integral)}

{\bf Notation}: 
$\int \int\int_E f(x,y,z)\, dV$  \quad \quad .

{\bf Inputs}: a {\bf function} $f(x,y,z)$ of three variables,
and a {\bf region} $E$ in space.

{\bf Output}: a {\bf number}.

{\bf How to Compute it}: 
Express the region $E$ either in $x-y-z$ order
or $x-z-y$ order etc., or in cylindrical, or in spherical
coordinaters.
Then convert the triple-integral into an {\bf iterated integral}
and do it {\bf one-step-at-a-time} from inside to outside.

{\bf Meanings}: 1. If $f(x,y,z)$ is positive, and you build
a {\bf four-dimensional}
building whose ``floor'' is the region $E$, and whose
ceiling-height above a typical point $(x,y,z)$ is given
by the function $f(x,y,z)$, then it is the 
{\bf four-dimensional ``volume''} of
that building.

2. If you have a solid body whose shape is $E$ and $f(x,y,z)$ is its
{\bf density function}, then it is the {\bf mass} of that solid-body.

{\bf Important Special case:} if $f(x,y,z)$ is $1$ then
you get the {\bf  volume} of $E$.

{\bf Concept 9: Gradient}

{\bf Notation}: $\g f(x,y,z)$ \quad \quad .

{\bf Inputs}: a {\bf function} $f(x,y,z)$ of three variables .

{\bf Output}: a {\bf vector-field}.

{\bf How to Compute it}: 
$\g f = \langle f_x, f_y, f_z \rangle$ \quad \quad .

{\bf Meaning}: the {\bf magnitude} of $\g f$
is the {\bf maximum rate of change} and its
direction is the {\bf direction where it changes the fastest}.


{\bf Concept 9: Curl}

{\bf Notation}: $curl \, {\bf F}$  (in math)
or $\g \times {\bf F}$ (in physics)

{\bf Input}: a {\bf vector-field} 
${\bf F}(x,y,z)$ of three components and three variables,
$\langle P,Q,R \rangle$.

{\bf Output}: another {\bf vector-field}.

{\bf How to Compute it}: 
$$
\g \times {\bf F}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
P & Q & R 
} \right \vert \quad .
$$

{\bf Meaning}: very important in Electricity and Magnetism.

{\bf Concept 10: Divergence}

{\bf Notation}: $div {\bf F}$  (in math)
or $\g \cdot {\bf F}$ (in physics) \quad \quad .

{\bf Input}: a {\bf vector-field} 
${\bf F}(x,y,z)$ of three components and three variables,
$\langle P,Q,R \rangle$.

{\bf Output}: a {\bf function} of $(x,y,z)$.

{\bf How to Compute it}: $P_x+Q_y+R_z$.

{\bf Meaning}: very important in Electricity and Magnetism.

{\bf Important relationships between concepts}

{\bf Green's Theorem} relates the notion of {\bf line-integral}
(of the vector-field kind over a {\bf closed} curve) in {\bf two} dimensions
with the notion of {\bf area-integral} (a.k.a. as {\bf double integral}).


{\bf Stokes's Theorem} relates the notion of {\bf line-integral}
(of the vector-field kind) in {\bf three} dimensions
with the notion of {\bf surface-integral} (of the vector-field kind)
over an {\bf open} surface [one that has a bounding-curve].

{\bf The Divergence Theorem} relates
the notion of {\bf surface-integral} (of the vector-field kind)
over a {\bf closed} surface with the notion of {\bf triple-integral}
(a.k.a. volume-integral) .


{\bf Important Simplifications}

1. For {\bf any} function $f(x,y,z)$ (that is ``nice'')
$curl \, ({\g f})= {\bf 0}$.

More explicitly: If you take any function $f(x,y,z)$ (you name it!)
and first take its {\bf gradient} getting some vector field
(a vector of functions). Then take the curl of that vector-field,
and {\bf surprise} you get the ${\bf 0}$ vector-field.


2. For {\bf any} vector-field ${\bf F}(x,y,z)$ (that is ``nice'')
$div \, (curl \, {\bf F})=0$.

More explicitly: If you take any vector-field ${\bf F}(x,y,z)$ (you name it!)
and first take its {\bf curl} getting some other vector field,
then take the {\bf divergence} of that vector-field,
and {\bf surprise} you get the $0$-function.

How to use these fact? Suppose that I give you
an extremely complicated function like
$$
f(x,y,z)=\cos(e^{\cos xyz + \sin z} + x+y+z^3 ) (x^3+7xyz-y^6)
$$
and ask you to compute $curl(\g f)$. If you do it stupidly, you
would first find $\g f$, and then take the curl of that.
This would take you two hours and you won't have time to do
anything else. But if you {\bf knew} the fact that
$curl\, (\g f)$ is {\bf always} ${\bf 0}$ it would take you one second.

{\bf Important shortcuts}

1. If you have an area-integral with the integrand being
a plain number of a region whose area you know from middle-school,
for example
$$
\int\int_D 5 \, dA \quad,
$$
You first take the 5 in front
$$
5 \int\int_D  \, dA \quad,
$$
and use the fact that when the integrand is $1$ the
{\bf area-integral} becomes plain {\bf area} so the
answer is $5$ times the area of $D$.

Remember: the area of a rectangle is: length times width;

the area of a triangle is: base times height over 2.

The area of a circle radius $r$ is $\pi r^2$.

The area of a semi-circle radius $r$ is $(\pi r^2)/2$.

2. Similarly if you have a {\bf surface-integral} with
the integrand being a number (remember the surface
area of a sphere radius $R$ is $4\pi R^2$,
that of an open hemisphere  half of that, that
of a closed cylinder radius $R$ height $h$ is
$2 \pi R h+ 2\pi r^2$) .


3. Similarly if you have a {\bf volume-integral} with
the integrand being a number (remember the 
volume of a sphere radius $R$ is $(4\pi/3)R^3$,
that of a hemisphere  half of that, that
of a cylinder radius $R$ height $h$ is
$ \pi R^2 h$) .

4. Similarly if you have a {\bf line-integral} 
(of the $ds$ kind)
with the integrand being a number (remember the length
(i.e. circumference) of a circle radius $R$ is
$2 \pi R$).


5. In {\bf polar} and {\bf cylindrical} coordinates
the ``phrase'' $x^2+y^2$ is $r^2$.

6. In {\bf spherical} coordinates
the ``phrase'' $x^2+y^2+z^2$ is $\rho^2$.


{\bf Use and Abuse of Nonsense Detection}

length, area, and volume can {\bf never} be negative numbers
but {\bf line-integrals}, {\bf area-integrals} and {\bf volume-integrals}
often are.

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