An interesting Curve

The area in question can be divided up into little rectangles and little triangles (all perpendicular to the ellipse) as shown.

Because the ellipse makes a full rotation, all the little triangles will add together to make a circle with radius 3. All the little rectangles will add together to make a rectangle with height $3$ and length $k$ where $k$ is the perimeter of the original ellipse. We can see that the total area of the shape in question is $9\pi+3k$ and the total perimeter is $6\pi+k$ . Now all we have to do is find $k$ . If a short line segment has horizontal width $dx$ and slope $m$ , the line segment length will be $\sqrt{1+m^2}dx$ . This means that $k$ will be equal to $2\int_{-5}^{5} \sqrt{1+f'^2}dx$ where $f(x)$ is the function of the ellipse. We can use an integral calculator to calculate $k=25.52699$ and $\text{Perimeter+Area}=4k+15\pi=\boxed{149.231}$

More generally, suppose we start with a curve $\alpha$ , and construct a curve $\beta$ that has a distance of $r$ from $\alpha$ . Then $\text{Length}(\beta) = \text{Length}(\alpha) + 2 \pi r$ and $\text{Area}(\beta) = \text{Area}(\alpha) + r \cdot \text{Length}(\alpha) + \pi r^2.$

See https://math.stackexchange.com/questions/58467/a-question-on-closed-convex-plane-curves-from-do-carmo . You can get some intuition for this result by considering the case where $\alpha$ is a polygon.

The parametric equation of the given ellipse is

$\mathbf{p}(t) = (5 \cos t, 3 \sin t)$

The tangent vector is given by

$\mathbf{p'}(t) = (-5 \sin t, 3 \cos t)$

Therefore, the unit normal is given by

$\mathbf{n}(t) = (3 \cos t, 5 \sin t) / \sqrt{ 9 \cos^2 t + 25 \sin^2 t }$

It follows that

$\mathbf{q}(t) = (5 \cos t, 3 \sin t) + 3 (3 \cos t, 5 \sin t) / \sqrt{ 9 \cos^2 t + 25 \sin^2 t }$

The derivative with respect to $t$ is given by

$\mathbf{q'}(t) = (-5 \sin t , 3 \cos t) + 3 (-3 \sin t, 5 \cos t) / \sqrt{9 \cos^2 t + 25 \sin^2 t} - 3 (3 \cos t, 5 \sin t) (-9 \cos t \sin t + 25 \sin t \cos t )/ (9 \cos^2 t + 25 \sin^2 t)^{(3/2)}$

$= (-5 \sin t, 3 \cos t) + 3 ( -3 \sin t, 5 \cos t) / \sqrt{ 9 \cos^2 t + 25 \sin^2 t} - 3 (3 \cos t, 5 \sin t) (8 \sin 2 t )/( 9 \cos^2 t + 25 \sin^2 t)^{(3/2)}$

To find the length of the curve, we integrate the length of $\mathbf{q'}$

$L = \displaystyle \int_{0}^{2 \pi} | \mathbf{q'}(t) | dt$

And the area enclosed by the curve is given by

$A = \frac{1}{2} | \displaystyle \int_{0}^{2 \pi} (\mathbf{q}_x \mathbf{q'}_y - \mathbf{q}_y \mathbf{q'}_x) dt |$

These integrations can be carried out very easily and accurately using Simpson's Rule numerical integration method, and the result is

$L =44.37655$ and $A = 151.97922$ , thus the area of the blue-shaded band is $A - \pi (5)(3) = 104.85533$ , and this makes the answer

$\lfloor 1000( 44.37655 + 104.85533) \rfloor = \lfloor 1000(149.23188) \rfloor = 149231$

Concentrating on the first quadrant, the ellipse is $y=3 \sqrt{1-\frac{x^{2}}{25}}$

The normal as a function of x is the opposite reciprocal of the derivative $N(x)=\sqrt{\frac{625}{9x^{2}}-\frac{25}{9}}$

For any $(x,y)$ on the ellipse, we can now add the correct amount of this normal to each coordinate to get a new point on the outer rim $(x+\frac{3}{\sqrt{1+N^{2}}},y+\frac{3N}{\sqrt{1+N^{2}}})$

Now let $x$ and $y$ each be mediated by parameter $a$ where $x=a+\frac{9a}{\sqrt{625-16a^{2}}}$ and $y=3\sqrt{1-\frac{a^{2}}{25}}(1+\frac{25}{\sqrt{625-16a^{2}}})$

This is where I leave out some gory details. From here I leaned heavily on WolframAlpha. Find $dx$ and $dy$ easily enough.

The area can now be found by $\int_{0}^{5} (dx\cdot y) da \approx 37.9948$ then subtract a quarter of the ellipse $\pi\cdot 5 \cdot 3 / 4$ to get $26.2138$

The perimeter is $\int_{0}^{5} \sqrt{dx^{2}+dy^{2}} da \approx 11.0941$

Multiply these by 4 and add them to get $S \approx 149.2317$ and so $\lfloor 1000S \rfloor =\boxed{149231}$

Both the quantities (area of the band and arc length of the outer curve) can be calculated by considering a small length $ds$ on the ellipse as shown below-

From the above image we can write down following equations-

$ds$ = $\sqrt{1 + y'^2}dx$

$r*d\theta$ = $ds$

r = $\frac{(1+y')^\frac{3}{2}}{y''}$

$dA$ = outer sector - inner sector = $\frac{(r+k)^2}{2}d\theta - \frac{r^2}{2}d\theta$ = $\frac{k^2}{2}d\theta - krd\theta$

$dL$ = $(1+\frac{r}{k})*ds$ (using similarity of both sectors)

With the help of above equations, we can eliminate $ds, r$ and $d\theta$ and express $dA$ and $dL$ just in terms of x. Integrating over one quadrant and using numerical method for the elliptic integral, we can arrive at the answer - 149.231

Taking inner ellipse as (5, 3) and 3 the thickness of the outer lane, we have x and y on the outer boundary as a function of angle 't' from x axis:

x(t)= 5cos(t)+9cos(t)/√f(t),

y(t)= 3sin(t) +15sin(t)/√f(t), where,

f(t)=(25cos(t)^2+9sin(t)^2)

Integrating by WolframAlpha a simplified expression of y(t){dx(t)/dt} dt from 0 to π/2 gives area as 37.99948. Subtracting 15π/4 and multiplying by 4 gives area of the outer lane of width 3 as 104.85533.

Similarly, integrating by WolframAlpha a simplified expression of √{(dx/dt)^2 + (dy/dt)^2}dt from 0 to π/2 and multiplying by 4 gives perimeter of the external boundary as 44.37655.

Sum of boundary area and perimeter is 104.85533+44.37655=149.23188, and Answer=149231.