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$T: \mathbb{R^3} \rightarrow \mathbb{R^3}$ is defined by $T(u,v,w) = (u \cos v \cos w, u \sin v \cos w, u \sin w)$. Show that $T$ is onto the unit sphere, $x^2 + y^2 + z^2 = 1$.

I believe I have to show:

$$ u \cos v \cos w = x \\ u \sin v \cos w = y \\ u \sin w = z $$

For $x^2 + y^2 + z^2 = 1$. Does that mean I have to solve for $u, v,$ and $w$? How can I do this? And how do I make my codomain restricted to the unit circle instead of $\mathbb{R^3}$?

user151682
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  • Your approach is both correct and incorrect. You're on the right track, but you're not thinking about what you actually have to do. It's not that you have to solve for $u,v,w$ but rather that you have to prove that the image of $T$ is a superset of the unit sphere. – Pockets May 19 '14 at 23:52
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    And in order to do that, just take $u=1.$ – mfl May 19 '14 at 23:53
  • What is the motivation for taking $u = 1$? – user151682 May 19 '14 at 23:54
  • @user151682: to have unit circle – mesel May 19 '14 at 23:56
  • @user151682, what your function $T$ is doing is generating points on a sphere of radius $u$, at an angle $v$ from the $x$-axis and $w$ from the $z$-axis. Your goal is to prove that the unit sphere is a subset of the image of $T$. What does restricting $u$ to 1 do? – Pockets May 20 '14 at 00:00
  • @SamuelLijin How do I prove that my function $T$ is generating points on a sphere of radius $u$ (at an angle $v$ from the $x$-axis and $w$ from the $z$-axis, etc.)? – user151682 May 20 '14 at 00:03
  • @ManuelFdzLpz you can also take $u = -1$ since there is a symmetry, you'll simply have different results for $v,w$ – DanZimm May 20 '14 at 00:07
  • You just have to use the fact $\forall \alpha : \sin^2 \alpha + \cos^2 \alpha = 1$ – NovaDenizen May 20 '14 at 00:40

1 Answers1

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Well first, the function is defined with a codomain of $\mathbb{R}^3$ so the question isn't asking to show that $T$ itself is onto, but that $\forall \; \vec{v} \in B_1(0) \; \exists \vec{w}$ s.t. $T(\vec{w}) = \vec{v}$ where $B_1(0)$ is the unit sphere.

In order to do this we need to take some arbitrary element of $B_1(0)$, meaning: $$ \text{Let } x,y,z \in \mathbb{R} \text{ s.t. } x^2 + y^2 + z^2 = 1 $$ and show that $\exists \; u,v,w$ s.t. $T(u,v,w) = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ If we solve the system of equations you have above: $$ u \cos v \cos w = x \\ u \sin v \cos w = y \\ u \sin w = z $$ with the property that $x^2 + y^2 + z^2 = 1$. First we can see that in order to have the property that $x^2 + y^2 + z^2 = 1$ we must have that $$ 1 = u^2 \cos^2 v \cos^2 w + u^2 \sin^2 v \cos^2 w + u^2 \sin^2 w = u^2 \left(\cos^2 w \left( \cos^2 v + \sin^2 v \right) + \sin^2 w \right) = u^2 $$ so that $u = \pm 1$. Continuing, for convenience let's choose $u = 1$ (note in general you cannot simply choose this, why?). So now we need to solve the system: $$ \cos v \cos w = x \\ \sin v \cos w = y \\ \sin w = z $$ Now we get from the last equation that $w = \arcsin z$ and find the following equations: $$ \cos v \cos \arcsin z = x \\ \sin v \cos \arcsin z = y $$ We can divide these equations to see that $$ \tan v = \frac{y}{x} $$ so that $v = \arctan \frac{y}{x}$. Now put $$ \vec{w} = \begin{bmatrix} 1 \\ \arctan y / x \\ \arcsin z \end{bmatrix} $$ so that $$ T(\vec{w}) = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$ And thus $T$ is onto the unit sphere, since $x,y,z$ were arbitrary. Hopefully this helps!

DanZimm
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