Coordinate conversion without graphing?

Question

Consider the region given by $R= \left\{x \ge 0, y \ge 0, z \ge 0, x^2+y^2+z^2 \le 1\right\}$

By letting $x = r\sin \phi \cos \theta, ~ y= r \sin \phi \sin \theta, ~ z= r\cos \phi$, show that

$$R = \left\{\theta \in [0, \pi/2], \phi \in [0, \pi/2], r \in [0,1]\right\}$$ (No graphing allowed. Determine $\theta, \phi, r$ from the parametric equations).

I think I see that $r \cos{\phi} \ge 0 \implies \phi \in [0,\pi/2].$ But sometimes this wouldn't be right because it would also depend on the second equation $x^2+y^2+z^2 \le 1$. I've seen that sometimes $\phi \in [0,\pi/4]$ but letting $r\cos \phi \ge 0$ would always give $\phi \in [0,\pi/2]$. And how do you find $\theta$?

Answer 1

Let $R=\{(x,y,z)\in \Bbb R^3: x,y,z\geq 0\land x^2+y^2+z^2\leq 1\},$ which is the proper way to write it. Let $S=\{(r\sin A\sin B, r\sin A\cos B, r\cos A): r\in [0,1]\land A,B\in [0,\pi /2]\}.$

Show (i) $S\subset R,$ and (ii) $R\subset S.$

(i). If $p=(x,y,z)=(r\sin A\sin B, r\sin A\cos B,r\cos B)=(x,y,z)\in S \;$ with $r\in [0,1]$ and $A,B\in [0,\pi /2]$ then $x,y,z\geq 0$, and by direct calculation $x^2+y^2+z^2=r^2\leq 1, $ so $p\in R$.

(ii). If $p=(x,y,z)\in R$ let $r=\sqrt {x^2+y^2+z^2}.$ Then $r\in [0,1].$

There exist $s,t\in [0,1]$ such that $s^2+t^2=1$ and $z=tr$ and $\sqrt {x^2+y^2}\;=sr.$ So there exists $B\in [0,\pi /2]$ such that $s=\sin B$ and $t=\cos B.$

There exist $u,v\in [0,1]$ such that $u^2+v^2=1$ and $x=u\sqrt {x^2+y^2}\;$ and $y=v\sqrt {x^2+y^2}.$ So there exists $A\in [0,\pi /2]$ such that $u=\sin A$ and $v=\cos A.$

Then $p=(x,y,z)=(r\sin A\sin B,r\sin A\cos B, r\cos B)\in S.$

Answer 2

$$\cos (\phi )\ge 0 \implies 0\le \phi \le \pi/2 $$ but depending on the problem you find the actual range of$ \phi$ which in this case is $$ 0\le \phi \le \pi/2$$

With regards to $\theta $ ,

Since both your $x$ and $y$ are non-negative, you have both $\sin(\theta)$ and $\cos (\theta)$ non-negative, thus $0\le \theta \le \pi /2 $

Answer 3

This is really a comment, not an answer, but it's too long. The point is that you can always tell what quadrant an angle lies in by the signs of the sine and cosine. This is clear from the polar form $(x,y)=(r\cos \theta, r\sin\theta)$.

In the first quadrant, both functions are positive. In the third quadrant, both are negative. In the second quadrant, sine is positive and cosine negative. In the fourth quadrant, sine is negative and cosine positive.

That is, cosine and sine have the same signs as $x$ and $y$, respectively.