Bigger than a circle

Calculus Level 4

The diagram above shows both a unit circle and a circle-like shape $S$ --slightly larger than the unit circle--defined by the equation $\sqrt{1-|x|} + \sqrt{1-|y|} = 1.$ What is the percentage by which the area of $S$ is bigger than the area of the unit circle?

Give your answer to 3 decimal places.

The answer is 6.1032954.

2 solutions

Arjen Vreugdenhil
Feb 8, 2017

Let's focus on the part of the shape in the first quadrant. We find the area by integration. First express $y$ in terms of $x$ : $y = 1 - (1 - \sqrt{1-x})^2 = 2\sqrt{1-x} - 1 + x.$ The area is $\int_0^1 y\,dx = \int_0^1 (2\sqrt{1-x} - 1 + x)\,dx = \left[-\frac 43 (1-x)^{3/2} - x + \frac 12 x^2\right]_0^1 = \frac43 - 1 + \frac12 = \frac56.$ This is one-quarter of the entire shape; the total area of shape $S$ is therefore $A_S = 4\times \frac56 = 3\tfrac13.$ Compare this to the area of the unit circle, $A_U = \pi$ : $\frac{A_S}{A_U} = \frac{3\tfrac13}\pi = 1.06103 \Longrightarrow \boxed{6.103}\%.$

The larger region closely resembles a superellipse . If you are interested, you can read more about it there. :)

Michael Huang - 4 years, 4 months ago

It is not a superellipse, but gets very close indeed. This inspired a new problem .

Arjen Vreugdenhil - 4 years, 4 months ago

This is strange, is there a reason why this curve looks so much like a circle?

By just looking at the first quadrant, the gradient at each point $(X,Y)$ can be found (via implicit differentiation ) to be $-\dfrac XY$ and $- \sqrt{ \dfrac{1-Y}{1-X} }$ for the circle and the non-circle, respectively. But I don't see how that helps....

Pi Han Goh - 4 years, 4 months ago

The non-circle runs through $(0,1), (1,0)$ , and $(0.75,0.75)$ . The circle runs through $(0,1), (1,0)$ , and $(\sqrt{\tfrac12}, \sqrt{\tfrac12}) \approx (0.71, 0.71)$ . That suggest some of the closeness.

Moreover, if we consider the polar representation $r(\theta)$ , we see that $r = 1$ for the circle, and $r$ varies between $1$ and $\tfrac34\sqrt2 \approx 1.061$ . This shows how close the two are.

Conjecture. For the non-circle, $r(\theta)$ is strictly increasing on $[0^\circ,45^\circ]$ , and strictly decreasing on $[45^\circ,90^\circ]$ .

Here is a start of a possible proof: first, from implicit differentiation we have $\sqrt{1-y}\:dx + \sqrt{1-x}\:dy = 0.$ From $x = r\cos\theta, y = r\sin\theta$ we find $dx = \frac x r\:dr - y\:d\theta;\ \ dy = \frac y r\:dr + x\:d\theta.$ Substitute these in the first equation to eliminate $dx$ and $dy$ : $(y\sqrt{1-x} + x\sqrt{1-y})\:\frac{dr}r = (y\sqrt{1-y} - x\sqrt{1-x})\:d\theta.$ If only we can prove that, for points on the non-circle in the first quadrant, $y\sqrt{1-y} > x\sqrt{1-x}$ iff $0 < y < x < 1$ we have proof.that $dr/d\theta > 0$ iff $x < y$ .

Arjen Vreugdenhil - 4 years, 4 months ago

1-|x|=sin⁴(t),1-|y|=cos⁴(t) |x|=1-sin⁴(t), |y|=1-cos⁴(t)

Nikola Djuric - 4 years, 3 months ago

x²+y²=(1-sin⁴(t))²+(1-cos⁴(t))²= 2-2(sin⁴(t)+cos⁴(t))+ sin^8(t)+cos^8(t)= 2-2(1-2sin²(t)cos²(t))+ (1-2sin²(t)cos²(t))²-2sin⁴tcos⁴t= 1+2sin⁴tcos⁴t= 1+sin⁴(2t)/8 So1≤x²+y²≤9/8 so we see it is very near x²+y²=1

Nikola Djuric - 4 years, 3 months ago

Ashish Gupta
Feb 12, 2017

Ah parametrization! This is probably the last thing that I would be thinking when solving this problem! Very creative.

Note that you can (slightly) simplify your integral $\displaystyle 4 \int_0^1 (1-t^2)(1-t^4) t \, dt$ by making the substitution $u =t^2$ .

Pi Han Goh - 4 years, 3 months ago

Bigger than a circle

The answer is 6.1032954.

2 solutions

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