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Youwen Wu 2025-03-03 00:43:59 -08:00
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@ -1106,5 +1106,63 @@ Any 3D shape can be built recursively of atomic objects.
Schley what are you doing???
== Signed area and volume
= Double integrals and some fun
We consider various types of integrals. Consult
#link("https://web.evanchen.cc/textbooks/poster-ints.pdf")[this chart] to
classify them.
Namely, we consider integrals of 0, 1, 2, and 3 dimensions, on scalar valued
function from $RR^0 -> RR$, $RR^1 -> RR$, $RR^2 -> RR$, and $RR^3 -> RR$.
To evaluate double integrals, we need to place notable importance on the
_region_ we integrate over.
== Integrating over rectangles
Integrating over a rectangle is super easy. It's like partial derivatives,
where we hold every other variable constant and integrate over our desired
value.
#example[
Evaluate
$ integral^6_0 integral^1_0 x y^2 dif x dif y $
We just crunch the numbers.
The first integral:
$
integral_0^1 x y^2 dif x = lr(1/2 x^2 y^2 |)_(x=0)^(x=1) = 1 / 2 y^2
$
The second integral:
$
integral_0^6 1 / 2 y^2 dif y = lr(1/6 y^3 |)_(y=0)^(y=6) = 36
$
]
== $x y$ integration without a rectangle
In general most 2D regions $cal(R)$ can still be done with $x y$ integration even when they aren't rectangles. In that case, we use the notation
$
integral.double_cal(R) f(x,y) dif x dif y := "integral of" f "over" cal(R)
$
If the region is given by inequalities, for instance, a unit disk, we would write
$
integral.double_(x^2 + y^2 <= 1) f(x,y) dif x dif y
$
== Swapping the order of integration
If our function is nice, then this is easy.
#theorem[Fubini's][
If the function that defines our surface is continuous, the double integral
over a rectangle can be evaluated in either order without a change to the
integral.
]
Otherwise, we want to swap the order of integration. We convert the limits of
integration back into inequality/region format, getting a region $cal(R)$ like
discussed in the previous section. Then evaluate that integral using standard
methods.

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@ -2261,7 +2261,7 @@ They are "easy" to use for finding the distributions of:
E[X^k] = mu_k, k = 1,2,...
$
Then the *moment generating function* of a random variable $X$ is defined by
$M_x(t) = E[e^(t x)]$, for the real variable $t$.
$M_x (t) = E[e^(t x)]$, for the real variable $t$.
]
All of the moments must be defined for the MGF to exist. The MGF looks like