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EDITED: Q11 is answered.

Please have a look at this origin post: Dog bone-shaped curve

In case not to make the question too long and make it horrific, I decided to open a new post for $2$ new subquestions,

11)(Answered, see comment by Yuriy S) Find the area (maybe not 'integral', since in definition to Riemann-integral, integral under x-axis is negative, but clearly area can't be negative) of the shaded green region and (*or) blue region, see the image below and here:enter image description here

The area of the dog bone, which is the region enclosed by curve $f$, is already integrated by Barry Cipra in the answer of the origin post, which gave a result $A≈3.52723$.

*Therefore, by the areas of circle and dog bone, I believe it is easy (Using simple $+,-,\times,\div$) to find the area of one of the $2$ regions whenever knowing the another region's area.

I believe one need to change the curve into functions to find the area, see the answers to the origin post, especially those by Barry Cipra and Yuriy S.

By the way, I think the description of the green shaded area is quite interesting, as is invented by myself:

$$\bigl(\left|x\cos a-y\sin a\right|^{x\cos a-y\sin a}>\left|x\sin a+y\cos a\right|^{x\sin a+y\cos a}\bigr)$$

$$\land\bigl(x\cdot\cos a-y\sin a<x\sin a+y\cos a\bigr)$$

$$\land\Biggl(\max\Bigl(\min(|x|^x-|y\|^y,-x+y),\min(-|x|^x+|y|^y,x-y)\Bigr)<0\Biggr)$$

(In this situation, $a=\frac{\pi}{2}$)

12) It will be fun if anyone can integrate both green and blue area in terms of $a$, while $a$ is the rotation degree (clockwise) of the red dog bone! (See the description of the green shaded area above)

See the image and here:enter image description hereWhat I discover so far, are, the area of the green region be $\lt \frac{A}{2}≈1.76$, while the area of the blue region should be $\lt\frac{A}{2}-\frac{\pi(\frac{2}{\sqrt e})^2}{2}\approx 1.338$ (As the radius of inscribed circle $=\frac{2}{\sqrt e}$), no matter what $a$ is :). Any help, such as verification of my discoveries right above, will be appreciate :).

Thank you for your answer(s).

Tony Ma
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    I have made that same picture as your first one. The green shape is approximated by an ellipse very well, using the known values of maxima obtained in some of the answers. See the blue ellipse here https://www.desmos.com/calculator/v9pctxqjii – Yuriy S Feb 19 '18 at 07:43
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    The blue region in your first picture has the area already known from the area of inscribed circle and the integral I from my second answer – Yuriy S Feb 19 '18 at 07:44
  • Oh sorry! I forget that you have integrated I! Than my first question is totally answered, I believe, there is no need for approximation by the ellipse. – Tony Ma Feb 19 '18 at 08:09
  • I'll use this opportunity to answer your question about parametric integral. For example, let's consider the case $x>y>0$, then $y=tx$, so we have: $$\int_0^1 y(t) x'(t)dt=\int_0^1t x(t) x'(t)dt= t x^2 \big|_0^1 -\int_0^1 (x^2 + t x x') dt$$ But the last integral is the same as the original one! So we get: $$2 \int_0^1t x(t) x'(t)dt = t x^2 \big|_0^1 -\int_0^1 x^2 dt $$ In our case the first part is zero, so we only need to consider the integral of $x^2$ – Yuriy S Feb 19 '18 at 09:37

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