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Calculate the area of the ellipse that you get when you rotate the ellipse $$\frac{x^2}{2}+y^2= 1$$ around the x-axis.

My approach has been to use the formula for rotation area from $-2$ to $2$. But this gives a complicated integral and I'm unsure about the limits $-2$ to $2$.

I would really appriciate a detailed solution to this

Mathias711
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iveqy
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  • http://math.stackexchange.com/questions/247332/area-of-an-ellipse – lab bhattacharjee May 13 '14 at 12:11
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    Wait, doesn't rotating an ellipse around an axis give you an ellipsoid in three dimensions such that this question should be asking you to compute the area of a surface? Rotating a curve around an axis doesn't seem to make much sense in two dimensions. – kahen May 13 '14 at 12:16

2 Answers2

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Draw a diagram.

Note that the maximum value of $x$ occurs when $y=0$, that is, when $x=\sqrt{2}$, and the minimum value of $x$ is $x=-\sqrt{2}$. The volume is $$\int_{-\sqrt{2}}^{\sqrt{2}} \pi y^2\,dx.$$ This is equal to $$\int_{-\sqrt{2}}^{\sqrt{2}} \pi \left(1-\frac{x^2}{2}\right)\,dx.$$

Remark: Because I like symmetry, and dislike negative numbers, I would actually integrate from $x=0$ to $x=\sqrt{2}$, and double the result.

André Nicolas
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Surface

$\frac{x^2}{2}+y^2=1$

$S=2\pi\int_{-a}^{a}yds=4\pi\int_{0}^{a}yds$

$y'= -\frac{x}{2y}$

$ds=\sqrt{1+y'^2}dx=\sqrt{1+\frac{x^2}{4y^2}}dx=\frac{\sqrt{4y^2+x^2}}{2y}dx$

$S=4\pi\int_{0}^{\sqrt{2}}y\frac{\sqrt{4y^2+x^2}}{2y}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4y^2+x^2}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4-x^2}dx$

Substitution: $x=2sint, dx=2cost\cdot dt$

$S=2\pi\int_{0}^{\pi/4}\sqrt{4-4sin^2t}\cdot 2costdt=8\pi\int_{0}^{\pi/4}cos^2tdt=$

$=4\pi\int_{0}^{\pi/4}(1+cos2t)dt=4\pi[t+\frac{sin2t}{2}]_{0}^{\pi/4}dt=\pi(\pi+2)$

georg
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