two point vertical, chose to do a cathedral, regretted it until I was done!

two point vertical, chose to do a cathedral, regretted it until I was done!

As I Live And Breathe (Logan x Veronica, PG-13)

*This is my first attempt at a Veronica Mars fan fiction, so any feedback on my characterization in particular would be wildly appreciated. Set immediately after The Wrath of Con in season one. Logan x Veronica fluff/comfort fic. Enjoy!*

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Veronica gasped out a laugh as the frigid water enveloped her fully, soaking her naked skin. Goosebumps erupted across her flesh and the taste of salt tingled on her lips. She dove under the water, heedless of her makeup and her curled hair, exalting in the feeling of the night’s diminished waves breaking against her as she came up for air.

The moon lit up the waves in a ghostly light, and she could barely make out the details of the limo waiting for her by the road. Seaweed tangled around her toes for a second before being swept away by the current. Maybe this wasn’t exactly the sexy, booze-fueled kind of skinny dipping Lilly would have had in mind for her first time… her best friend would have had it as a salacious rendezvous with some attractive, older guy or a drunken interlude to a party between the two of them; one where they could share secrets and talk of boys and the plans they could share for the future.

A future Lilly wasn’t ever going to have…

Her fingers found the pendent hanging from her neck, wrapping around it, and she breathed a plume of misty air into the sky. Maybe this wasn’t what Lilly would have planned for Veronica’s introduction to the scandalous world of skinny dipping, but maybe her best friend would be proud that Veronica was finally crossing something off her “Never Have I Ever” list.

“Veronica Mars?” Veronica spun in the water, sending an arc of water flying around her as her arms instinctively came up to wrap around her chest despite the depth. “As I live and breathe.”

“*Logan?!”*

Conceptualising Integration

*Cont’d from “Visualising integration of 3D Cartesian-based volumes”, see “Visualising integration of 2D Cartesian-based areas”*

This post is best viewed on my blog, click here to be taken to its permalink.

I’m sure we know how to go about integrating a circle whose centre point is placed on the origin – particularly in polar co-ordinates – but what about one displaced from the origin in Cartesian co-ordinates? Not such a simple problem.

We’ll start by visualising the scenario geometrically. Consider a circle of radius *r* displaced from the origin by a distance *R*. We’ll say for simplicity that *R* acts purely in the *x*-direction, meaning it has no vertical component. Hence, the scenario is pictured as below.

There are multiple ways we can approach this problem but we’ll go for simplicity whilst maintaining some challenge, of course. Let’s start by finding a function to represent the circle’s boundaries, so to obtain a contour (well, it’s not *strictly* a contour but I’m gonna call it that, sorry) with which to integrate over. It should be clear that, as it is, the circle cannot be represented as a function since it would require two *y*-values for one *x*-value. However, if we manipulate the scenario a bit it becomes a more viable option.

Now we have a semi-circle! As you can see, no two points intersect the same vertical line. Hence, a function *y*(*x*) can be found to represent the bounding contour line.

By trigonometry, we know how to express the radius of a circle as a product of its *x* and *y* components. In the most general sense, for a circle with its origin at the co-ordinate (*a*, *b*) this is,

r^{2}=(x–a)^{2}+ (y–b)^{2}

The proof for this is simply Pythagoras’ theorem so I will leave it out. Furthermore, in our case,

r^{2}= (x–R)^{2}+y^{2}

which is represented by the red line in the following diagram.

Hence, our function can be expressed as

y(x) = (r^{2}– (x–R)^{2})^{½}

Now we can use this to set the integration limits. The shape
is bounded in the x-axis by *R* – *r* and *R* + *r*.
Therefore our *x* limits are

R–r<x<R+r

In the *y*-direction,
however, the shape is bounded by the red line shown above which is given by our
function *y*(*x*). Hence, we can say that the limits in *y* are,

0 <

y<y(x)

Since this notation is a little confusing, more explicitly our limits are,

0 <

y< (r^{2}– (x–R)^{2})^{½}

Using these limits, we can express the area integral, ½ A, in
one of two ways: Either as the integral of an infinitesimal area element *dA* = *dxdy* over the
limits given above (discussed in detail in a previous post), or as the integral
of the function *y*(*x*) between the limits of *x*. To be explicit about this, we’ll use
the former since it is the most generally useful.

Hence the area of our semi-circle is,

½

A= ∯_{y}_{(}_{lim}_{ x}_{)}dA

which essentially tells us that half
the area of our circle is given by the area bounded by the function *y*(*x*),
where *y*(*x*) is bounded by the limits of *x*.
This is unconventional notation (and mathematicians would be tearing their hair out at such misuse of notation) but it is satisfactory for our purposes. Substituting the Cartesian
area element *dxdy* and applying the limits outlined by
*y*(*x*)
we get

A= 2 ∫_{R}_{-r}^{R}^{+r}∫_{0}^{√(}^{r}^{² – (x – R)²)}dydx

Since this series is all about visualising and conceptualising the processes taking place during integration, what does this look integration look like?

This animation shows how each infinitesimal element plays a part in the integration process to find the area. First the *dy *element scans the *y*-axis to outline the function y(x), which the *dx *element is dragged along and bounded by.

We can now evaluate the integral directly. Note that we cannot
separate the double integral since the limits of *y* are dependent on *x*.

A= 2 ∫_{R}_{-r}^{R}^{+r}[^{}y]|_{0}^{√(r² – (x – R)²)}dx

A= 2 ∫_{R}_{-r}^{R}^{+r}[(^{}r^{2}– (x–R)^{2})^{½}- 0]dx

A= 2 ∫_{R}_{-r}^{R}^{+r}(^{}r^{2}– (x–R)^{2})^{½}dx

This is where the integral gets a little more complicated.

Using substitution, we can let

u(x) =x–R⇒ du = dx

and evaluate the limits for this substitution,

u(x→R+r) = (R+r) –R= +r

u(x→R–r) = (R–r) –R= –r

which allows our integral to become

A= 2 ∫_{-r}^{+r}(^{}r^{2}–u^{2})^{½}du

A further substitution is required. With manipulation, we could
use the trigonometric identity cos²(*x*)
+ sin²(*x*) = 1 to our advantage.
Hence, we shall make the substitution,

u(x) =rsin [v(u)]

which implies,

/^{du}=_{dv}r⋅/^{d}[sin (_{dv}v)] ⇒du=rcos (v)dv

Again, we can evaluate the limits of *v*(*u*).

v(u) = sin^{-1}(^{u}^{(x)}/)_{r}

v(u→ +r) = sin^{-1}(/^{r}) = sin_{r}^{-1}(1) =^{π}/_{2}

v(u→ –r) = sin^{-1}(–/^{r}) = sin_{r}^{-1}(– 1) = –^{π}/_{2}

which makes the integral

A= 2 ∫_{-π/2}^{ +π/2 }(r^{2}–r^{2 }sin^{2}v)^{1}^{/2}rcosvdv

Looks way more complicated, right? Let’s make some simplifications. Start by taking out the constants,

A= 2r∫_{-π/2}^{ +π/2 }(r^{2}–r^{2 }sin^{2}v)^{1}^{/2}cosvdv

and factoring out common factor of *r*².

A= 2r∫_{-π/2}^{ +π/2 }[r^{2}(1 –^{}sin^{2}v)]^{1}^{/2}cosvdv

Now, remember that trig. identity I
mentioned earlier? Let’s apply it here, wherein 1 –^{}sin² *v*) = cos² *v*.

A= 2r∫_{-π/2}^{ +π/2 }[r^{2}cos^{2}v]^{1}^{/2}cosvdv

Cancel out the square with the root and take out the resulting constant.

A= 2r∫_{-π/2}^{ +π/2 }rcosvcosvdv

A= 2r^{2}∫_{-π/2}^{ +π/2 }cos^{2}vdv

Nearly there! Here, we can use another trigonometric
identity: The half-angle identity, which states that cos² *v* = ^{1}/_{2} + ^{(cos 2v)}/_{2}.

A= 2r^{2}∫_{-π/2}^{ +π/2 }[^{1}/_{2 }+^{(cos 2v)}/_{2}]dv

Take out the constant and separate out the integrals since ∫
[*f*(*x*) + *g*(*x*)] *dx*
= ∫ *f*(*x*) *dx* + ∫ *g*(*x*)
dx.

A= 2 (^{1}/_{2})r^{2}{ ∫_{-π/2}^{ +π/2 }(1)dv_{}+ ∫_{-π/2}^{ +π/2}cos 2vdv}

Now, we can evaluate the first integral with ease.

A=r^{2}{ [v]|_{-π/2}^{ +π/2 }+ ∫_{-π/2}^{ +π/2}cos 2vdv}

A=r^{2}{ [^{π}/_{2}– (–^{π}/_{2})] + ∫_{-π/2}^{ +π/2}cos 2vdv}

A=r^{2}{ [^{π}/_{2}+^{π}/_{2})] + ∫_{-π/2}^{ +π/2}cos 2vdv}

A=r^{2}{ π + ∫_{-π/2}^{ +π/2}cos 2vdv}

Unfortunately, the second integral isn’t so easy. We have to make yet another substitution.

w(v) = 2v⇒dv=/^{dw}_{2}

and evaluate the limits.

w(v→ +^{π}/_{2}) = 2(^{π}/_{2}) = + π

w(v→ −^{ π}/_{2}) = 2(−^{π}/_{2}) = − π

Hence, the integral is

A=r^{2}{ π + ½ ∫_{-π}^{ +π}coswdw}

which evaluates to

A=r^{2}{ π + ½ [sinw]|_{-π}^{ +π}}

A=r^{2}{ π + ½ [sin (π) – sin (– π)]}

A=r^{2}{ π + ½ [0 – 0]}

Finally, we find that the area of our circle is

A= πr^{2}

Revolutionary stuff, eh? (pardon the pun)

Although the result is something we already knew, the proof is still essential and the process used to find it is valuable knowledge since we can now apply this to a much less general case. In addition, it is useful to know how to integrate more complicated objects than cubes and rectangles in Cartesian co-ordinates.

I will use and expand upon this result to integrate a torus (i.e. a ring-doughnut shape), following the challenge informally set by tumblr user voidpuzzle.

Two Great Mistakes

**Fandom: **Marvel

Request:
*Can
you please do an AU where the reader is finds out their pregnant with
lots of fluff and cute Bucky moments! PLUS CUTE BELLY MOMENTS please!!
***Word Count:** 1072**Warnings:** none

**Authors note: **This is queued as I am working today and tomorrow (The whole weekend in my country) So i will probably post the next request in a few days :)

The
white and pink plastic stick rested in my hand as the two stripes
pointed vertically. “Crap” I mumble as panic and anxiety lingers at the
back of my throat like a bad aftertaste.

Tears well up in my eyes and
threaten to fall. I breathe slowly and chuck away the packaging and test
underneath the layers of unhygienic rubbish in the bin so no one will
know. “Of course they will know, you are growing a goddamn baby inside
you, they’ll know” I say quietly to myself.

colors!!