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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.

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.

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

Under a smooth locally invertible coordinate map $\Phi$ into Euclidean space, the positive volume element in a multiple integral is multiplied by the absolute Jacobian determinant. In three-dimensional spherical coordinates, $dV=r^2\sin \theta drd\theta d\phi$; in cylindrical coordinates, $dV=rdrd\theta dz$. The absolute value distinguishes positive integration measure from a signed orientation convention.