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[?] Calculate this surface integral?

Posted 4 years ago

I want to compute the surface area of a spheroid. If I define the spheroid as an implicit region in 3 dimensions and integrate it numerically I get the correct answer. But if I define the spheroid as an ellipse of rotation and integrate then I am not getting the same result. Doing the integral one way

re = ImplicitRegion[(x^2 + y^2)/4 + z^2 == 1.0, {x, y, z}];
NIntegrate[1, {x, y, z} \[Element] re]

gives the correct result.

But if I define the region as an ellipse

re2 = ImplicitRegion[(x^2/4 + z^2 == 1.0) && z >= 0, {x, z}];

and integrate the function 2 pi z to account for the rotation about the x axis over that region, the result of the integration

2 NIntegrate[2 \[Pi] z, {x, z} \[Element] re2]

is 42.9569.

I get a different result if, rather than defining an implicit region, I just do an integral of 2 pi z times a differential line length along the ellipse.

y[x_] := Sqrt[1 - (x/2)^2]
yp[x_] := x/(2 Sqrt[2^2 - x^2])
2 NIntegrate[ 2 \[Pi] y[x] Sqrt[1 + yp[x]^2], {x, 0, 2}]

In this case the answer is 21.4784, half of the prior result.

What is wrong with my thinking about calculating the surface area? And why do the last two calculations that I thought compute the same thing result in a factor of 2 difference?

POSTED BY: Mike Luntz
2 Replies

Wrong axis for the rotational symmetry. Change restriction to x>=0 and integrand to 2*Pi*x and it will give the expected result.

POSTED BY: Daniel Lichtblau
Posted 4 years ago

Thanks Daniel. That was really dumb of me. I should have recognized my mistake and would have if I had plotted the region.

POSTED BY: Mike Luntz
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