I would like to calculate the angle, $\theta$ , between the x-axis and the lines tangent to this axisymmetric paraboloid at any point x, along the x-axis. The body is currently defined by a radial distribution as a function of x by the equation: $R(x)=0.2*x^{(1/2)}$, but I am unsure how to transform this to Cartesian coordinates (which I think would put me on the right track). Please see diagram of the problem below.
How do I do I find the angle between the body and the x-axis for any point x?
If I understood correctly, you’re trying to find the angle between the $x$-axis and the tangent line to the surface that lies in the plane defined by the $x$-axis and the point of tangency. The intersection of any half-plane bounded by the $x$-axis with this surface is just a copy of the radial distribution curve. Therefore, $\tan\theta = R'(x)$.
We can also go the long way around. An implicit equation of the surface is $R(x)^2=y^2+z^2$, with gradient $(2R(x)R'(x),-2y,-2z)$. A normal to the plane that contains the point $(x,y,z)$ and the $x$-axis is $(0,-z,y)$, therefore the direction of the line of intersection of this plane with the tangent plane to the surface at this point is $$(R(x)R'(x),-y,-z)\times(0,-z,y) = \left(y^2+z^2,R(x)R'(x)y,R(x)R'(x)z\right).$$ Since $y^2+z^2=R(x)^2$, this is equal to $$R(x) \left(R(x),R'(x)y,R'(x)z\right).$$ Assuming $R(x)\gt0$, by taking the dot product of this with $(1,0,0)$, we have $$\cos\theta = {R(x)\over\sqrt{R(x)^2+R'(x)^2(y^2+z^2)}} = \frac1{\sqrt{1+R'(x)^2}}.$$ This is consistent with $\tan\theta=R'(x)$.
If you’re looking for the angle between any tangent line and the $x$-axis, you can use the normal to the surface derived above. The tangent lines are all perpendicular to this vector, and once you have a direction for the line, you can use the dot product with $(1,0,0)$ to compute the angle, as above.