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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 cylindrical coordinates. The real square root restricts the projection to $0\le r\le 1$, and there $r-1\le 0\le \sqrt{1-r^2}$. Since $\nabla x={\mathbf{e}}_x$, the volume integral isFor an independent boundary calculation, split $\partial V$ into an upper hemisphere $S_{+}$ and a lower right circular conical surface $S_{-}$. They meet in the circle $r=1,z=0$ and there is no additional planar face.

On $S_{+}$, use the parametrizationThe outward unit normal vector is $\mathbf{R}$, and the surface area element is $\sin \alpha d\alpha d\theta$. The $y$ and $z$ components of ${\iint }_{S_{+}}x\mathbf{n}dS$ vanish by the angular integrals of $\sin \theta \cos \theta$ and $\cos \theta$. The $x$ component isOn $S_{-}$, the parametrization $(r\cos \theta ,r\sin \theta ,r-1)$ has outward vector area elementIts negative vertical component points out below the solid. Multiplying by $x=r\cos \theta$ and using the same angular symmetry in integration givesTherefore the independently evaluated surface integral isin agreement with the volume integral. The isolated apex of the cone does not contribute to the surface integral.

For a volume source of constant mass density $\rho$, use $dm=\rho d{V}^{\prime }$. Its Newtonian gravitational potential is the volume integralThe source and observation position vectors are ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$, respectively; $G$ is the gravitational constant. With this sign convention positive source mass gives negative Newtonian gravitational potential.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

Let $V$ be a bounded region with piecewise smooth closed boundary $\partial V$, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus the volume integral of the divergence equals the outward boundary flux integral. For a region containing a hole, the outward unit normal vector on its inner boundary points into that hole.

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.

Express the position vector in the suggested orthonormal basis as $\mathbf{r}=X{\mathbf{e}}_1+Y{\mathbf{e}}_2+Z{\mathbf{e}}_3$. ThenThis orthogonal transformation preserves $x^2+y^2+z^2=X^2+Y^2+Z^2$ and has Jacobian determinant of absolute value one, so $dV=dXdYdZ$. The basis actually reverses orientation; the absolute value of the Jacobian determinant is why no minus sign enters the volume integral. Also $x-y=\sqrt{2}Z$, so the half-space becomes $Z\ge 1/2$.

At each fixed $Z\in [1/2,1]$, the section is the disk $X^2+Y^2\le 1-Z^2$, of area $\pi (1-Z^2)$. Its reflection symmetry in $X$ makes the integral of $X$ vanish. Hence the required triple integral isThe negative sign is also consistent with the cap lying in the direction $(1,-1,0)$, where its average $y$ coordinate is negative.

For each of the three constant Cartesian unit vectors ${\mathbf{e}}_j$, apply the divergence theorem to the vector field $\varphi {\mathbf{e}}_j$. Its divergence is ${\partial }_j\varphi$, soThese are precisely the three components ofThus the volume integral of the gradient is the boundary surface integral weighted by the vector surface element $d\mathbf{S}=\mathbf{n}dS$. The regularity needed here is that $\varphi$ be continuously differentiable on a neighbourhood of the closure of $R$.

Use cylindrical coordinates $(\rho ,\theta ,z)$. The right circular cone is described by $0\le z\le 1$ and $0\le \rho \le a(1-z)$, and the divergence of $\mathbf{F}=(0,0,z^2)$ is $2z$. Direct integration gives

To check the boundary flux integral independently, first consider the base $z=0$. Its outward unit normal vector is $-\mathbf{k}$, but $\mathbf{F}=0$ there, so its flux is zero. Parametrize the lateral surface byThe cross productpoints outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral isAdding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

The mass density becomes $\rho =\mu (2+\cos \theta )$. Integrate it over the sphere using the spherical polar coordinates and volume element above:The angular integral is $4$, since ${\int }_0^{\pi }\cos \theta \sin \theta d\theta =0$. ThereforeThe term proportional to $z/r$ contributes zero by reflection across the equatorial plane. Thus the answer is also the volume $4\pi a^3/3$ times the average mass density $2\mu$. The undefined direction at the origin affects a single point and has no effect on this volume integral.

The divergence is $\nabla \cdot F=2xyz+2xyz+2xyz=6xyz$. With outward orientation, the divergence theorem converts the surface flux into a volume integral:Directly, the three faces lying in coordinate planes contribute zero, and each opposite face contributes $1/4$, confirming the sign and normalization.

The ellipsoid and the exponential factor are invariant under separate reflections of any coordinate. Split the integrand into its two polynomial terms. The term $x^4{y}^{2}z$ times that exponential is odd in $z$, so its volume integral vanishes by reflecting $z\mapsto -z$. The other term $x{y}^2{z}^4$ times the exponential is odd in $x$, so its volume integral vanishes by reflecting $x\mapsto -x$. Both integrands are continuous on a bounded region. Therefore odd-integrand cancellation by reflection gives

Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

The height is constant along each circular ring, whose area element after angular integration is $2\pi rdr$. The volume integral in cylindrical coordinates is thereforeHere $u=r/R$. Using integration by parts, ${\int }_0^{1}u{e}^{-u}du=[-(u+1)e^{-u}{]}_0^1=1-2/e$. HenceThe dimensionless coefficient is positive, and the cube root of a volume has the required physical dimension of length.

Integrate by horizontal slices. At a fixed height $0\le z\le H$, the parabolic boundary gives $|x|\le L\sqrt{z/H}$; the other inequality gives $|y|\le b(2-z/H)$. Thus the cross-section is a rectangle of areaThe volume integral, followed by integration by substitution $u=z/H$, isThis has units of volume. As a check on the geometry, the width decreases with height while the length of the horizontal slice increases from zero to $2L$.

In the $x,z$ plane, the embankment lies above the parabola $z=H{x}^2/L^2$ and below $z=H$. The bore has circular cross-sectionIt touches the valley floor at the origin. To check that this entire disk lies inside the projected embankment, its equation gives $x^2\le Dz-z^2\le Dz$. Since $HD/L^2<1$, it follows that $H{x}^2/L^2\le z$, with strict inequality away from the bottom contact. Also $z\le D<H$. Thus no circular segment needs to be subtracted from the tunnel section.

For each $(x,z)$ in the disk, the length of bore inside the material is the local width $2b(2-z/H)$. The required volume integral is thereforeThe disk has area $A=\pi D^2/4$ and centroid height $D/2$, so ${\iint }_{\mathcal{D}}zdxdz=AD/2$. Applying the volume of a solid with affine thickness formula givesThe width varies along the circular section; multiplying its area by the top width alone would miss this variation.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.

Let $V$ be the enclosed region with piecewise smooth closed boundary $S$, oriented by its outward unit normal $\mathbf{n}$. For a continuously differentiable vector field on a neighbourhood of $V$, the divergence theorem statesThe outward flux integral equals the volume integral of the divergence.

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore