# Find limits of integration for the interior region of sphere with center $(a,0,0)$ and radius $a$ using spherical coordinates

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.

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

$(x-a)^2 + y^2 + z^2 = a^2\\ x^2 + y^2 + z$ = 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$a.$Cylindrical. $$x = x+a\\ y = r \sin(\theta)\\ z = r \cos(\theta)$$ x from$-\sqrt {a^2-r^2}$to$\sqrt {a^2-r^2}$or, $$x = x\\ y = r \sin(\theta)\\ z = r \cos(\theta)$$ x from$a-\sqrt {a^2-r^2}$to$a+\sqrt {a^2-r^2}\$