Tricky Surface Parametrization

Question

I am to parametrize the surface given by the ellipse $$9(z-1)^2 + x^2 = 1$$ in the $xz$-plane and rotated about the $x$-axis. I then have to find the volume of the region enclosed.

The concept of "rotated about the $x$-axis" is causing me some difficulty.

I have come up with $$x = cos\theta$$$$z = (\frac1 3 sin\theta + 1)sin\phi$$

Which I am not even sure is right, and then the best I can get for $y$ is $$y = zcos\phi$$

There is a 3d render of it https://i.stack.imgur.com/rsWxJ.png

Any help would be appreciated.

bubba · Answer 1 · 2014-03-21T06:17:57.250

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Hint: if you take the point $(a,0,b)$ lying in the $xz$-plane, and you rotate it around the $z$-axis by an angle $\phi$, you get the point $(x,y,z) = (a\cos\phi, a\sin\phi, b)$. If you draw a picture, this should be clear. Or, if you like algebra more than geometry/pictures, just observe that such points $(x,y,z)$ satisfy $x^2 + y^2 = a^2$, so they lie on a circle of radius $a$ whose center is on the $z$-axis.

Edit:
Since the hint didn't work ... we put $a=\cos\theta$ and $b = 1 + \tfrac13 \sin\theta$, and we get the parameterization:

$$ x = \cos\theta \cos\phi \\ y = \cos\theta \sin\phi \\ z = 1 + \tfrac13 \sin\theta $$

edited Mar 21 '14 at 06:17

answered Mar 21 '14 at 00:45

bubba

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So the parametrization should be $$x = cos\theta$$ $$y = zsin\phi$$ $$z = (\frac13 sin\theta +1)cos\phi$$? – Vivid Mar 21 '14 at 03:42
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Nope. See additions to my answer. – bubba Mar 21 '14 at 06:18

score 0 · Answer 2 · answered Mar 21 '14 at 05:24

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Area of ellipse: $ \pi a b = \pi/3 $, by Pappus's Centroid Theorem , the centroid is moving $2 \pi $ under rotation , so the Volume would be $2 \pi^2 / 3$

answered Mar 21 '14 at 05:24

Alan

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Tricky Surface Parametrization

2 Answers2