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Writing $r^2=x^2+y^2+z^2$, the gradient is $\mathbf{v}=2e^{r^2}(x,y,z)$. Its divergence, equivalently the Laplacian of $\varphi$, isOn the unit sphere this scalar is the constant $10e$. The lower hemisphere has surface area $2\pi$, soThis is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

Integrate the normal component on each face directly, with outward normals. Each unlisted pair of coordinates ranges independently from zero to one:For example, on $x=1$ the integral is ${\int }_0^1{\int }_0^{1}2(y+z)dydz=2$. Adding all six contributions givesThis calculation uses the six explicit surface integrals, independently of the divergence theorem.

Separate the closed cylinder into its curved side and its two end disks, all with outward normal vectors. On the side, $\rho =R$ and $\mathbf{n}=\hat{\mathbf{\rho }}$, soAt the top, $z=h/2$ and $\mathbf{n}=\mathbf{k}$, giving $(\nabla \times \mathbf{v})\cdot \mathbf{n}=\alpha h$. At the bottom, both $z$ and the axial normal vector change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi \alpha R^{2}h$. Adding all three surface integrals,This explicit cancellation is consistent with the fact that the divergence of a curl is zero. The two disk contributions add rather than cancel.

For the outward orientation of a spherical cap, its vector area element isIn the surface integral defining the vector area, the $x$ and $y$ components vanish by integration over a complete azimuth. The remaining component isConsequentlyThis also equals the area of the cap's projection onto the horizontal plane, directed upwards.

Each of the three surface pieces is invariant under $y\mapsto -y$, while $y^3$ is an odd function under that reflection. Its surface integral therefore vanishes on each piece separately. Only $z$ contributes.

On the cylindrical side, use cylindrical coordinates $x=a\cos \varphi$, $y=a\sin \varphi$, with area element $dS=ad\varphi dz$:The upper disc has constant $z=b$ and area $\pi a^2$, giving $I_{\mathrm{top}}=\pi a^{2}b$. On the hemispherical indentation, use spherical coordinates with $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$:Adding the three contributions yieldsThis is a scalar surface integral, so the indentation contributes positively through $dS$; an inward or outward orientation does not change its sign.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

On the sphere of radius $a$, the outward unit normal is $\mathbf{n}=(x,y,z)/a$. Thus $\mathbf{F}\cdot \mathbf{n}=a$ everywhere. Multiplying by the surface area $4\pi a^2$ givesThe same surface integral follows in spherical coordinates from $dS=a^2\sin \vartheta d\vartheta d\phi$.

On the disc $z=0$, the upward normal selects only the third component, which equals $e^0/(1+0)=1$. Hence the surface integral is the area of the unit disk:

For $k=1$, the vector field is $(x^{2}y,x{y}^2,z{)}^T$. On the cone,The first term has zero integral by symmetry. Using polar coordinates for the remaining double integral givesOn the paraboloid $z={\rho }^2-1$,Again the $xy$ term cancels by symmetry, givingTherefore direct surface integral yieldsNo separate disk at $z=0$ is included: that rim joins the two surfaces and is not an additional face of the closed boundary.

Both patches project onto the unit disk $D$ in the $xy$-plane. Their outward oriented surface elements areThe cone's apex has zero area and does not affect the surface integral. For $k=0$, the vector field is $(y,x,1{)}^T$, soThe terms involving $xy$ vanish by reflection symmetry of the disk. Thus the total outward flux isThis calculation evaluates the two surface integrals directly.

The cube is $1\le x\le 2$, $0\le y,z\le 1$, so $|x|=x$ throughout it. Evaluate the six outward surface integrals directly.

On $x=2$ the normal component is $2$, giving flux $2$; on $x=1$ it is $-1$, giving flux $-1$. On $y=1$ the normal component is $2x$, giving ${\int }_1^2{\int }_0^{1}2xdzdx=3$; on $y=0$ it is zero. On $z=1$ the normal component is $e^{-1}$ and on $z=0$ it is $-1$, so those unit-area faces give $e^{-1}$ and $-1$.

Adding all six contributions yieldsOnly direct surface integral was used, as requested.

On the unit sphere, the outward unit normal is $\mathbf{n}=(x,y,z{)}^T$, so the normal component of the radial vector field is $\mathbf{F}\cdot \mathbf{n}=x^2+y^2+z^2=1$. The direct surface integral isHere $dS=\sin \theta d\theta d\varphi$ is the unit-sphere area element of a surface in spherical coordinates.

The curl of a gradient is zero pointwise, so its surface integral over this sphere isFor this particular polynomial, $\nabla \Phi =2(x,y,z{)}^T$, whose curl vanishes directly as well.

The divergence of the position vector field is $\nabla \cdot \mathbf{r}=3$. At height $z$, the horizontal cross-section is a disk of area $\pi (1-z^2)$. By the divergence theorem, the outward flux integral is three times the enclosed volume:ThereforeA direct check separates the two pieces of the surface integral. On the unit sphere, the outward unit normal is $\mathbf{r}$, so $\mathbf{r}\cdot \mathbf{n}=1$ and the curved contribution is $2\pi (z_0+1)$. On the top disk, $\mathbf{n}={\mathbf{e}}_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same flux.

Use the two pieces of the surface integral directly; this avoids introducing an unnecessary second derivative of $f$. On the unit sphere, $\mathbf{n}=\mathbf{r}$ and $\mathbf{F}\cdot \mathbf{n}=f^{\prime }(z)$. A spherical band between heights $z$ and $z+dz$ has area $2\pi dz$, so the curved flux isby the fundamental theorem of calculus. On the horizontal top disk, $\mathbf{F}\cdot \mathbf{n}=z_0{f}^{\prime }(z_0)$ and the area is $\pi (1-z_0^2)$. ThusAs usual for these surface integrals, the first derivative is assumed sufficiently integrable for the displayed fundamental theorem of calculus evaluation; a continuous first derivative suffices. Mere pointwise differentiability without such an integrability assumption need not make an ordinary flux integral exist.

Using Cartesian coordinates, the Levi-Civita symbol and the Einstein summation convention, the constant-vector assumption givesRecognizing the divergence and directional derivative provesHere the contractions use the Kronecker delta.

For an oriented piecewise smooth surface $S$ with boundary $C$, Stokes theorem stateswhere the orientations obey the right-hand rule and $\mathbf{Q}$ is continuously differentiable near $S$. Apply it to $\mathbf{Q}=\mathbf{F}\times \mathbf{a}$. The scalar triple product givesThe last integrand is $\mathbf{a}\cdot [(d\mathbf{S}\times \nabla )\times \mathbf{F}]$, as the component formulashows. Since $\mathbf{a}$ is arbitrary, the cross-product form of Stokes theorem follows:The derivatives in the surface expression act on $\mathbf{F}$, not on the surface element.

For the specified square, the path is clockwise when viewed from positive $z$, so the compatible unit normal is $-{\mathbf{e}}_z$. On the first and fourth edges, $d\mathbf{l}$ and $\mathbf{r}$ are parallel, giving zero cross product. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf{l}\times \mathbf{r}={\mathbf{e}}_{z}dx$ contributes ${\mathbf{e}}_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-{\mathbf{e}}_z{\int }_1^{0}dy={\mathbf{e}}_z$. Thus the line integral is $2{\mathbf{e}}_z$.

For an independent surface integral, ${\partial }_i{F}_j={\delta }_{ij}$ and $\nabla \cdot \mathbf{F}=3$ for the position vector field. Its integrand is therefore $d\mathbf{S}-3d\mathbf{S}=-2d\mathbf{S}$. The square has unit area and downward orientation, giving

Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

In cylindrical coordinates, the surface has radius $r=\sqrt{2}z$. It is the lateral surface of a right circular cone, with vertex at the origin and axis along positive $z$. Its height is $h$ and the limiting upper rim has radius $\sqrt{2}h$; the horizontal disk at that rim is absent. The half-angle from the vertical satisfiesThe upper rim is excluded by $z<h$, which does not affect any convergent surface integral. The vertex likewise contributes no area.

First compute the curl directly:Reflection symmetry in $x$ makes the last component integrate to zero. The middle component gives the hemisphere's volume, soFor an independent surface integral, the closed boundary consists of the curved hemisphere and its flat base. The right-hand side in part (b) is ${\int }_S\mathbf{n}\times \mathbf{F}dS$.

On the curved part use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. There,The last component has zero angular integral. The first two components giveOn the base, $z=0$ and the outward normal vector is $-{\mathbf{e}}_z$. Thus $\mathbf{n}\times \mathbf{F}=(x^2+y^2,0,0)$. In polar coordinates its integral isThe first components cancel, and therefore both required sides equal

Since $\mathbf{F}=\nabla \Omega$, its divergence is the Laplacian of $\Omega$:The divergence theorem changes the outward flux through the whole closed boundary into $I={\int }_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the Jacobian determinant from part (c),Both the curved face and the top disk are included in this surface integral through the divergence theorem.