I am asked to find limits of integration for the interior region of sphere with center $(a,0,0)$ and radius $a$ using spherical coordinates. How can one do that?

I know that one may use

$$ x = r \cos(\theta) \sin(\phi)\\ y = r \sin(\theta) \sin(\phi)\\ z = r \cos(\phi) $$

Is it possible do to the same with cylindrical coordinates?

Thank you.

  • $\begingroup$ What if you recenter the origin? Just rename the x-coordinate, and redefine the function to be integrated in the new coordinates? $\endgroup$ – vinnief May 20 '16 at 1:18

you have a few choices.


$(x-a)^2 + y^2 + z^2 = a^2\\ x^2 + y^2 + z$ = 2ax$

Spherical... since x is the "special one", I would suggest.

$$x = r \cos(\phi)\\ y = r \sin(\theta) \sin(\phi)\\ z = r \cos(\theta) \sin(\phi)$$

Plug these into your equation for the sphere and, $r^2 = 2a\,r\cos\phi $
$r$ will range from $0$ to $2a\cos\phi, \theta$ from $0$ to $2\pi, \phi$ from $0$ to $\pi/2$

If you went with the traditional.

$$x =r \cos(\theta) \sin(\phi)\\ y = r \sin(\theta) \sin(\phi)\\ z = r \cos(\phi)$$

Then $r$ will range from $0$ to $2a\cos\theta\sin\phi, \theta$ from$-\pi/2$ to $\pi/2$

how about...Taking the traditional and translating it.

$$x =r \cos(\theta) \sin(\phi) + a\\ y = r \sin(\theta) \sin(\phi)\\ z = r \cos(\phi)$$

And $r$ goes from $0$ to $a.$


$$x = x+a\\ y = r \sin(\theta)\\ z = r \cos(\theta)$$

x from $-\sqrt {a^2-r^2}$ to $\sqrt {a^2-r^2}$


$$x = x\\ y = r \sin(\theta)\\ z = r \cos(\theta)$$

x from $a-\sqrt {a^2-r^2}$ to $a+\sqrt {a^2-r^2}$



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