# An example of divergence in spherical coordinates

I've found the following example in a vector calculus book: the divergence of the vector field $$\vec F(x,y,z) = x\vec i + y\vec j - z \vec k$$ in spherical coordinates is

$$\nabla \cdot \vec F(\rho,\phi,\theta) = -3\cos(2\phi).$$ I understood all the passages of the book and I've found the same result. But why the following argument does not holds true: in cartesian coordinates $$\nabla \cdot \vec F(x,y,z)=1$$ and then $$\nabla \cdot \vec F(\rho,\phi,\theta)=1$$? In my mind if $$\nabla \vec G(x,y,z)=xy$$ (for instance) then $$\nabla \vec G(\rho,\phi,\theta) = \rho^2\sin\phi^2 \sin \theta \cos\theta$$. Am I missing something?

• Can you show the passages in the book? Because your argument is valid and it's not clear how they got $-3\cos 2\phi$ – Vasily Mitch Feb 1 at 17:15
• Ok, the idea is to use the formula for divergence in the spherical coordinates. The orthogonal basis in spherical coordinates is $T_\rho = (\sin \phi \cos\theta,\sin\phi\sin\theta,\cos\phi), T_\phi = (\cos\phi\cos\theta, \cos\phi\sin\theta,-\cos\phi), T_\theta = (-\sin\phi\sin\theta,\sin\phi\cos\theta,0)$. We have to write $\vec F = F_1 T_\rho + F_2 T_\phi + F_3+T_\theta$. Notice that $\vec F(x,y,z) = \rho\sin\phi\cos\theta \vec i + \rho\sin\phi\sin\theta \vec j + \rho\cos\phi \vec k$. So are able to compute $F_1 = \vec F \cdot \vec T_\rho$ (analogously $F_2$ and $F_3$). (continue) – user85353 Feb 1 at 17:24
• These calculations leads to: $F_1=-\rho\cos(2\phi), F_2=F_3=0$. Now we put directly in the formula of divergence and we get the answer. – user85353 Feb 1 at 17:26
• Another example of the book calculates the Laplacian in spherical coordinates of the function $f(x,y,z) = x^2+y^2-z^2$. The book says that the answer isn't $1$.. for me the same argument can be used. I'm so confused... – user85353 Feb 1 at 17:30
• Could you show a scan of the calculations in the book? – md2perpe Feb 1 at 17:36