Perhaps something along these lines:

With[{r = Sqrt[x^2 + y^2], \[Phi] = ArcTan[x, y]}, 
 Show[ContourPlot3D[
   Evaluate[{r^2 + z^2 - 1 == 0, r == Simplify[Sin[\[Phi]]]}], {x, -1,
     1}, {y, -1, 1}, {z, -1, 1}, ContourStyle -> Opacity[0.3], 
   AxesLabel -> {"x-axis", "y-axis", "z-axis"}, MeshFunctions -> {}, 
   LabelStyle -> Directive[Bold, Medium, Red]],
  VectorPlot3D[{Grad[r^2 + z^2 - 1, {x, y, z}], 
    Grad[r - Simplify@Sin[\[Phi]], {x, y, z}]}, {x, -1, 1}, {y, -1, 
    1}, {z, -1, 1}]]]

That worked very nicely, thank you. I also added axes - makes it easy to keep track when rotating the image....

I also wanted Opacity, but despite being in the documentation, the following failed :

With[{r = Sqrt[x^2 + y^2], \[Phi] = ArcTan[x, y]}, 
 ContourPlot3D[{r^2 + z^2 - 1 == 0, r == Sin[\[Phi]]}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1},
  PlotStyle -> Opacity[0.5], 
  AxesLabel -> {"x-axis", "y-axis", "z-axis"}, 
  LabelStyle -> Directive[Bold, Medium, Red]]]

'Unknown option name has been used" - any idas how to make the functions partly opaque? Thanks

Thanks Sander, that is nice. ( Delay because I had lots of hassles upgrading to win 10, eventually had to rebuild my PC)

Is there a command to Plot this without the contour lines? I tried replacing with GraphPlot3D but this wont accept some of the options ....

Nice work, but you can try something along these lines:

With[{r = Sqrt[x^2 + y^2], \[Phi] = ArcTan[x, y]}, 
 ContourPlot3D[{r^2 + z^2 - 1 == 0, r == Sin[\[Phi]]}, {x, -1, 
   1}, {y, -1, 1}, {z, -3, 3}]]

Hi - one of the 2 eqtns I want to plot is {r = Sin[[Phi]]} But for coords - it's more that the other function I wish to plot simultaneously is a cylinder. So is the conclusion I would have to convert any cylindrical functions into Cartesian to plot them?

You want to plot a sphere using cylindrical co-ordinates? $x= r\cos(\phi), y=r \sin(\phi), z = z$, so $x^2+y^2=r^2$ and your equation defines a sphere circle in the plane at height $z$. If that really must be done, then $r$ grows from 0 (at $z=-1$) to 1 (at z = 0) and decreases back to 0 (at $z=1$) and of course this is given by $r=\sqrt(1-z^2)$. So one types (because ParametricPlot3D[] draws into an euclidean orthonormal co-ordiante system) just

ParametricPlot3D[{Sqrt[1 - z^2] Cos[\[Phi]], Sqrt[1 - z^2] Sin[\[Phi]], z}, {z, -1, 1}, {\[Phi], 0, 2 \[Pi]}]

to see as expected

That's more complicated than I expected. Surely Mathematica can directly use cylindrical (or polar) cords?

The documentation says to use SetCoordinates[Cylindrical] - but I don't get the expected output from that, which should be Cylindrical[Rr, Ttheta, Zz] - and we should in theory then be able to directly do what I want ...

But I tried the above, then ContourPlot3D[ Rr^2 + Zz^2 - 1 == 0, {Rr, -5, 5}, {Ttheta, 0, 2 [Pi]}, {Zz, -1, 1}] - and got the same warped 'thing'. Is there a bug with SetCoordinates perhaps?

I'm not sure what you're trying to do there but I'd guess that if you have in mind a discrete range of values for Phi eg

phi = {0,Pi/6,Pi/4,Pi/3,Pi/2,Pi,2Pi/3,3Pi/4,5Pi/6,Pi,7Pi/6,5Pi/4,4Pi/3,2Pi}

then you'd probably be able to apply appropriate transformations to the bounds of the variables inside

ContourPlot

What I have in mind is something like

rLoBound[x_,phi_]:=(appropriate calcs)
rHiBound[x_,phi_]:=(appropriate calcs)
zLoBound[x_,phi_]:=(appropriate calcs)
zHiBound[x_,phi_]:=(appropriate calcs)

ContourPlot3D[{r^2 + z^2 - 1 == 0}, {r, rLoBound[-5,#], rHiBound[5,#]}, {z, zLoBound[-3,#], zHiBound[3,#]}]&/@phi