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Parametrize the graph by $\mathbf{r}(x,y)=(x,y,f(x,y))$ over its projection $D$ onto the $xy$ plane. The tangent vectors are $(1,0,f_x)$ and $(0,1,f_y)$, so their cross product is $(-f_x,-f_y,1)$. Hence the scalar surface element isandThe printed subscript $S$ is shorthand for integrating over the projected parameter region: $dxdy$ belongs to $D$, while the square-root factor converts projected area into actual surface area. The corresponding upward vector surface element of a graph is $(-f_x,-f_y,1)dxdy$.

The divergence theorem states that a continuously differentiable vector field on a bounded region with piecewise smooth boundary satisfieswith the unit normal pointing out of $V$. Apply it separately to $\mathbf{u}=f{\mathbf{e}}_i$, where ${\mathbf{e}}_i$ is each constant Cartesian basis vector. Since $\nabla \cdot (f{\mathbf{e}}_i)={\partial }_{i}f$ and $(f{\mathbf{e}}_i)\cdot \mathbf{n}=f{n}_i$, the three component identities combine to give

For the specified scalar field, $\nabla f=(z,0,x)$. The $x$ term integrates to zero by reflection symmetry. At height $z$, the region has a disk cross-section of area $\pi (a^2-cz)$, and $0\le z\le a^2/c$. Therefore the volume integral isTo check the surface integral independently, split the boundary into the base disk and the curved graph $z=g(x,y)=(a^2-x^2-y^2)/c$. On the base $f=0$, so its contribution vanishes. The region lies below the curved graph, whose outward vector surface element of a graph isOver the disk $x^2+y^2\le a^2$, the $y$ and $z$ components of $fd\mathbf{S}$ are odd in $x$ and integrate to zero. Its $x$ component gives, in polar coordinates,Thus the two explicitly evaluated integrals agree, with the outward orientation included.

Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt{2}\le z\le \sqrt{5}$, so its projection is the annulus $1\le r\le 2$. The surface is the corresponding band of the upper sheet of a two-sheeted hyperboloid, bounded by circles of radii $1$ and $2$ at heights $\sqrt{2}$ and $\sqrt{5}$.

For the parametrization $\mathbf{R}(x,y)=(x,y,g(x,y))$, its tangent vectors are $(1,0,x/z)$ and $(0,1,y/z)$. Their cross product gives the upward vector surface element of a graph:Its magnitude gives the scalar surface area of a graph:Both formulas apply on $1\le \sqrt{x^2+y^2}\le 2$. Reversing the orientation changes only the sign of the vector element.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

In cylindrical coordinates, $\rho =(x^2+y^2{)}^{1/2}$ and $\mathbf{F}=(f(\rho )x/\rho ,f(\rho )y/\rho ,0)$. The chain rule gives ${\partial }_x\rho =x/\rho$ and ${\partial }_y\rho =y/\rho$. ThereforeThis is cylindrically radial divergence, valid for $\rho >0$.

The bounded domain is outside the unit cylinder, below the inverse-square surface and above the lower plane:The upper plane touches only the rim $\rho =1,z=1$; it is not an extra positive-area top face. Revolving the shaded meridional region below around the $z$ axis gives the three-dimensional domain.

With the Jacobian determinant $dV=\rho d\rho d\varphi dz$ of cylindrical coordinates,Zero divergence gives $(\rho f{)}^{\prime }=0$, hence the most general solution on the positive radial interval isAlthough this vector field is singular on the axis if $A\ne 0$, the entire closure of $V$ lies at $\rho \ge 1$, so it is smooth on a neighbourhood of $V$ and the divergence theorem applies.

Its volume integral of divergence is zero. Compute the outward surface integral directly. On the inner cylinder, $\mathbf{n}=-{\mathbf{e}}_{\rho }$, $dS=d\varphi dz$, and the flux is $-2\pi A(1-1/4)=-3\pi A/2$. On the curved roof $z=g(\rho )={\rho }^{-2}$, the outward vector surface element of a graph isTherefore its flux isThe horizontal bottom has zero flux and the top rim has zero surface area. Summing giveswhich verifies the divergence theorem with every boundary contribution accounted for.