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For radial mass density $\rho (r)=A{r}^{-2}(1-e^{-r/r_0})$ within radius $R>0$, integration against the spherical volume element gives $M=4\pi ARf(r_0/R)$. The dimensionless quantity $q=r_0/R>0$ hasThus $f$ is positive, strictly decreasing and convex, with $f(0^{+})=1$ and $f(\infty )=0$. Its asymptotic forms are $f(q)=1-q+q{e}^{-1/q}$ near zero and $f(q)=1/(2q)-1/(6q^2)+O(q^{-3})$ at infinity. The density's central $1/r$ divergence is integrable against the factor $r^{2}dr$.

For a spherical surface of radius $R$ with exterior mass density $\rho (r)={\rho }_0{e}^{-(r-R)/h_0}$, integrate the spherical volume element to obtain $M=4\pi {\rho }_0{\int }_0^{\infty }(R+h{)}^2{e}^{-h/h_0}dh$. The moments ${\int }_0^{\infty }{h}^n{e}^{-h/h_0}dh=n!h_0^{n+1}$ give the displayed expression. Its leading thin-atmosphere approximation is surface area times column mass, $4\pi R^2{\rho }_0{h}_0$; the two positive correction terms account for the growth of shell area with altitude.

For general coordinates $q=(a,b,c)$ expressed as functions of $u=(u,v,w)$, the chain rule gives $dq=Jdu$, whereThe Jacobian determinant is the oriented volume scaling of this linearized map. Thus the unsigned volume element transforms as $dadbdc=|\det J|dudvdw$. Where the coordinate change is invertible, the inverse Jacobian determinant is $1/\det J$, by the Jacobian chain rule.

For cylindrical coordinates, use $x=r\cos \varphi$, $y=r\sin \varphi$, and unchanged height $z$. Direct differentiation giveswhose determinant is $r$. ThereforeBoth directions are displayed because “from Cartesian to cylindrical” can refer either to the derivative of the new coordinates or to the factor used in transforming a Cartesian volume integral. They are reciprocals; the cylindrical axis is a coordinate degeneracy.

For a bounded volume $V$ with outward-oriented piecewise smooth boundary $S$, the divergence theorem statesHere $G$ is a continuously differentiable vector field on a neighborhood of $V$, $n$ the outward unit normal vector, and $dV$ the volume element. The left side is its outward flux.

For a twice continuously differentiable field $F$, equality of mixed partial derivatives givesApply the divergence theorem to $G=\nabla \times F$ to obtainThe regularity hypotheses justify differentiating and integrating the curl; a field singularity cannot simply be ignored in a volume argument.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Take $0<r_1<r_2$, so the shell excludes the possible singularity at the origin. The divergence and curl of a radial vector field formula in spherical coordinates givesWith volume element $dV=r^2\sin \theta drd\theta d\varphi$, direct integration givesFor the independent boundary calculation, the outer sphere has outward unit normal vector $+\hat{\mathbf{r}}$ and the inner sphere has outward unit normal vector $-\hat{\mathbf{r}}$. Their fluxes areTheir sum equals the direct volume integral, verifying the divergence theorem. The restriction $\beta \ne 2$ permits the displayed power-antiderivative step. At $\beta =2$, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.

Use the spherical polar coordinates volume element $dV=r^2\sin \theta drd\theta d\varphi$. Since $1+\cos \theta \ge 0$ throughout $0\le \theta \le \pi$, the solid has radial range $0\le r\le 1+\cos \theta$ and full azimuthal range $0\le \varphi <2\pi$. Its volume isPut $u=1+\cos \theta$, so $du=-\sin \theta d\theta$ and the endpoints become $2$ and $0$. ConsequentlyThe polar equation describes the entire three-dimensional surface, so the spherical polar coordinates volume element, rather than a two-dimensional polar area element, is required.

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.

Use spherical polar coordinates with $\theta$ measured from the positive $z$ axis and $\varphi$ around that axis:The standard full-space ranges are $r\ge 0$, $0\le \theta \le \pi$, $0\le \varphi <2\pi$. The factor $r^2\sin \theta$ is the absolute Jacobian determinant and gives the volume element.

In cylindrical coordinates the spherical cap has $1<z<\sqrt{2}$, $0<r<\sqrt{2-z^2}$ and $0<\theta <2\pi$. Its mass density is $kz$ and the volume element is $rdrd\theta dz$. ThereforeThis is the mass of a spherical cap with height-proportional density with $R=\sqrt{2}$ and $c=1$.

For its curved surface area, write $z=\sqrt{2-r^2}$ with $0\le r\le 1$. The surface integral of one becomesIt agrees with the curved area of a spherical cap, $2\pi Rh$, for $h=\sqrt{2}-1$. The planar closing disk is not part of the requested curved surface.

In cylindrical coordinates, the two surfaces meet when $9-r^2=r^2$, giving $r=3/\sqrt{2}$ and $z=9/2$. The solid has $0\le r\le 3/\sqrt{2}$, $0\le \theta \le 2\pi$, and $r^2\le z\le 9-r^2$. With volume element $rdzdrd\theta$,The upper-minus-lower height is nonnegative throughout that radial interval.

The divergence is $y^2+x^2+z^2=r^2$. In spherical coordinates, the angular integral is $4\pi$ and the volume element contributes $r^{2}dr$. Thus

The volume element in cylindrical coordinates is $rdzdrd\theta$. Integrating from the base to the curved surface givesHenceFor a complete rotation this is half the volume of the enclosing cylinder.

In spherical coordinates, $\theta$ is the polar angle from the positive $z$ axis and $\varphi$ the azimuthal angle. The Jacobian determinant of the coordinate transformation gives the volume element

In spherical coordinates, integrating over the angular variables gives the volume element $4\pi r^{2}dr$ for a spherical shell. The mass density therefore gives the total massThus

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 volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.

In spherical coordinates, with polar angle $0\le \theta \le \pi$ and azimuth $0\le \phi <2\pi$, the volume element isThe mass density is radial, so integrating over both angles givesWriting $q=r_0/R>0$, this becomesAlthough the mass density diverges like $A/(r_{0}r)$ near the origin, multiplication by the spherical volume element gives an integrable expression. Thus the central singularity does not produce infinite mass.