Let $C$ be the cylinder $x^2+y^2=1$, let $f\colon C \rightarrow\mathbb R^3$ be the function $f(x,y,z) = (x\cos z, y\cos z,\sin z)$. Prove that the image of $f$ is precisely the unit sphere $S^2$.
Image of a function on a surface
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differential-geometry
vector-spaces
2 Answers
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FIRST PART: showing that the image of $C$ under $f$ is a subset of $S^2$:
$(x\cos(z))^2+(y\cos(z))^2+(\sin(z))^2=x^2 \cos^2(z)+ y^2\cos^2(z) + \sin^2(z)= $
$=\cos^2(z)(x^2+y^2)+\sin^2(z) = ... = 1 $ because $x^2 + y^2 = 1 $ since we're on the cylinder. That is, the image points lie on $S^2$.
SECOND PART: I leave this to you. You have to show that every point of $S^2$ is an image point.
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2@ninja: set up the three equations, $a=x\cos{z}$, $b=y\cos{z}$, $c=\sin{z}$. All you have to do is show that you can solve for $x$, $y$ and $z$. You can find $z$ right away. Then you can find $x$ and $y$. Finally check that $x^2+y^2=1$. – 2011-11-04
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$(xcosz)^2+(ycosz)^2+(sinz)^2= (x^2+y^2)(cosz)^2+(sinz)^2=1$. So any point in the image of $f(x,y,z)$ is on $S^2$.
We also need to show that any point on $S^2$ is in the image of $f$, but this is trivial to check.