How does one find the area of this?

Hello everyone! The relation of the image above is x2+y2=2yx^2+y^2 = 2^y. How do you find the area of the egg shape?

*Note: There is a section above the droplet, which should not be accounted for.

Made with Desmos graphing calculator

Note by Timothy Cao
1 year, 10 months ago

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A different question about this graph: Consider graphs of the form x2+y2=ayx^2+y^2=a^y. If a=2a=2, the graph is as above. If a=2.09a=2.09, the egg shape is now connected to the upper segment of the graph but if a=2.08a=2.08, the egg shape is still distinct. Therefore, at what value of aa does this egg first connect to the upper part of the graph?

Kaan Berk Erdogmus - 1 year, 9 months ago

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Actually, that question has a precise answer: a=e2/e2.0871a = e^{2/e} \approx 2.0871.

x2+y2=ay.x^2 + y^2 = a^y.

At the transition point, the tip of the "egg" and the dip in the rest of the graph meet each other; due to symmetry, this must happen somewhere on the line x=0x = 0, and the connection will have a vertical tangent. Thus we set dx=0dx = 0 in the total derivative and substitute ay=y2a^y = y^2:

2xdx+2ydy=(lna)aydy2x\:dx + 2y\:dy = (\ln a) a^y\:dy 2y=(lna)ay=(lna)y2.2y = (\ln a) a^y = (\ln a)y^2. lna=2y      a=e2/y.\ln a = \frac 2 y\ \ \ \therefore\ \ \ a = e^{2/y}.

Now substitute x=0x = 0 and this result in the original equation: y2=ay=(e2/y)y=e2;y^2 = a^y = (e^{2/y})^y = e^2; y=e, a=e2/e.y = e,\ a = e^{2/e}.

Arjen Vreugdenhil - 1 year, 9 months ago

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You can also generalize this solution for graphs of the form x2n+y2n=ayx^{2n}+y^{2n} = a^y. The general solution for the critical aa is a=e2nea = e^\frac{2n}{e}

Kaan Berk Erdogmus - 1 year, 9 months ago

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Ayush Jain - 1 year, 9 months ago

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Ayush Jain - 1 year, 9 months ago

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Ayush Jain - 1 year, 9 months ago

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Ayush Jain - 1 year, 9 months ago

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Ayush Jain - 1 year, 9 months ago

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It is approximately 2.316, but I think you will have to resort to numerical integration as I did. Formally, we have A=c222yy2dy,A = \int_{-c}^2 2\sqrt{2^y - y^2}dy, where c0.767>0c \approx 0.767 > 0 is such that (12)c=c2(\tfrac12)^c = c^2; however, there is no hope to solve this integral algebraically.

Arjen Vreugdenhil - 1 year, 10 months ago

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