(a) Find a vector $V(x,y,z)$ normal to the surface
$$z=\sqrt{x^2+y^2}+(x^2+y^2)^{3/2}$$
at a general point $(x,y,z)$ of the surface, $(x,y,z)\neq (0,0,0)$.
(b) Find the cosine of the angle $\theta$ between $V(x,y,z)$ and the z-axis and determine the limit of $\operatorname{cos}\theta$ as $(x,y,z)\to (0,0,0)$.
The answer for these are (a) $(1+3x^2+3y^2)(x\mathbf{i} +y\mathbf{j})-(x^2+y^2)^{1/2}\mathbf{k}$ or any scalar multiple thereof.
(b) $\operatorname{cos}\theta =-[1+(1+3x^2+3y^2)^2]^{-1/2}$; $cos\theta \to -\frac{1}{2}\sqrt{2}$ as $(x,y,z)\to (0,0,0)$.
First I don't see how the answer for (a) turns out to be as such. My idea is to set the equation of the surface to be $f(x,y,z)=\sqrt{x^2+y^2}+(x^2+y^2)^{3/2}-z=0$, and get the gradient of $f$, which should be normal to the surface, however, this does not match the one given in the answer.
Also, taking the answer for (a) as given, I tried to find the cosine of $\theta$ by the equation $cos\theta=\frac{a\dot b}{|a||b|}$, however, this turns out to be \frac{-(x^2+y^2)^{1/2}}{[1+(1+3x^2+3y^2)^2]^{-1/2}, which is again not the answer.
I would greatly appreciate it if anyone could explain what's wrong about my answers and how I can get the right ones.
Well, I have tried doing (b) for you but I am totally lost in the massive algebraic fuster cluck involved. However, for (a), I assume when you took the gradient of $$ f(x,y,z)=\sqrt{x^2+y^2}+(x^2+y^2)^{3/2}-z, $$ you got $$ \nabla f=[x(x^2+y^2)^{-1/2}+3x(x^2+y^2)^{1/2}]\vec{i}+[y(x^2+y^2)^{-1/2}+3y(x^2+y^2)^{1/2}]\vec{j}-\vec{k}. $$
You only need to multiply this vector field with $(x^2+y^2)^{1/2}$, and perform some basic algebraic manipulations and you'll get the supposed answer.