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Limits of Surface Integral in Polar Coordinates

An object occupies the region V:x2+y2≤9,z∈[−1,1] V: x^2+y^2 \leq 9, z \in[-1,1]

I have to find the surface integral ∬sF⋅nds \iint_{s} \mathbf{F} \cdot \mathbf{n} \mathrm{d}s , where F=(x(z+1),−1,(z+1)y2) \mathbf{F} = (x(z+1),-1,(z+1)y^2)
Splitting the surface into the three regions of the cylinder, the top and bottom are fine, but I can't understand why certain limits were used when trying to find the surface integral for z∈(0,1) z \in (0,1).

Using cylindrical polar coordinates:

F⋅nds=[9cos2(ϕ)(z+1)−3sin(ϕ)]dϕdz \mathbf{F} \cdot \mathbf{n} ds = [9cos^2(\phi)(z+1) -3sin(\phi)]\mathrm{d}\phi \mathrm{d}z

Now I understand that the limits of ϕ\phi are 0 and2π 2\pi. But I can't see what the limits of r are. The solutions says its -1 and 1, but I can't see why.
Sure you mean r and not z? Note that you're given that z∈[−1,1]z \in [-1, 1].
Original post by DFranklin
Sure you mean r and not z? Note that you're given that z∈[−1,1]z \in [-1, 1].


Right, yes. Just a careless error then.

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