In SI units, Gauss's law for a closed surface with outward unit normal vector isSpherical symmetry and the absence of an imposed external field make the electric field radial, $\mathbf{E}=E_r(r){\mathbf{e}}_r$. For $0\le r\le a$ (with the physical radius $a>0$),Outside the charge distribution this becomes $Q(a)=8\pi K{a}^5/15$. Applying Gauss's law on spheres givesBoth formulas agree at $r=a$, and the first tends to zero at the origin. Thus the total charge is $8\pi K{a}^5/15$. Negative $K$ reverses the field direction. If the radius is zero the distribution vanishes; a negative value of $a$ gives no nonnegative-radius support under the printed piecewise inequalities and likewise gives the zero distribution.

For a holonomic conservative system the Lagrangian is $L(q,\dot{q},t)=T-V$, the difference between kinetic energy and potential energy. The Euler-Lagrange equations areIf $\partial L/\partial q_j=0$, this gives $d{p}_j/dt=0$ for the generalized momentum $p_j=\partial L/\partial {\dot{q}}_j$. Such a coordinate is a cyclic coordinate.

In cylindrical coordinates the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. The holonomic constraint $z=r^2/a$ gives $\dot{z}=2r\dot{r}/a$, while gravity has potential energy $V=mgz$. ThereforeThere is no explicit $\theta$ dependence, so $\theta$ is a cyclic coordinate andThis is angular momentum about the vertical symmetry axis. At $r=0$ the polar angle is singular, but the Cartesian angular momentum $m(x\dot{y}-y\dot{x})$ remains well defined and has the same conserved value.

Let $\eta$ be a smooth variation with $\eta (a)=\eta (b)=0$. Assuming the usual continuous second derivatives of $F$, stationarity of the functional under $y+t\eta$ gives its first variationIntegration by parts removes the derivative of $\eta$; its boundary term vanishes because the endpoint values are fixed. Thusfor every such variation. The fundamental lemma of the calculus of variations implies the Euler-Lagrange equation $F_y=(d/dx)F_{y^{\prime }}$.

For arc length, $F=\sqrt{1+(y^{\prime }{)}^2}$, soThe function $s\mapsto s/\sqrt{1+s^2}$ is strictly increasing; hence $y^{\prime }$ is constant. With endpoints $(a,A)$ and $(b,B)$, $a<b$, the extremal isIt is also the global shortest curve: integrating the unit displacement vector dotted with the curve velocity gives the endpoint distance, which is at most the integrated speed by Cauchy-Schwarz inequality; the straight segment attains equality. A vertical segment is described parametrically or after rotating axes, since it cannot be represented by $y(x)$ on a nontrivial horizontal interval.

A topological space is connected when it cannot be written as the union of two disjoint nonempty open sets. Equivalently, the only subsets that are simultaneously open and closed are the empty set and the whole space. For a subset, connectedness refers to its subspace topology.

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.

The annulus is open and hence Lebesgue measurable; multiplying its indicator by the continuous Gaussian gives a measurable function. It is bounded by the indicator of $[-b,b{]}^2$, whose Lebesgue integral is $4b^2$. Thus it is Lebesgue integrable by comparison.

Use the change of variables formula with polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Removing one radial cut makes this a smooth one-to-one coordinate map; the cut is a null set and does not affect the integral. The absolute Jacobian determinant is $r$. By Fubini theorem,The antiderivative is $-e^{-r^2/2}$, which also makes the sign of the answer transparent.

Use the agreed text encoding to split the plaintext into integer blocks $m_j\in \{0,\dots ,1072\}$. RSA encrypts each block by modular exponentiation:Because $1073=29\cdot 37$ is a product of distinct primes, its Euler totient function is $\phi (1073)=28\cdot 36=1008$. The inverse exponent follows fromThus a decryption exponent is 275, and the three recovered blocks areTo check every block, including blocks not coprime to the modulus, reduce modulo each prime. If $m_j$ is nonzero modulo 29, Fermat's little theorem gives $m_j^{3025}=m_j$ because $28\mid 3024$; if it is zero the equality is immediate. The same argument works modulo 37 because $36\mid 3024$. The Chinese remainder theorem then gives the equality modulo 1073. Finally reverse the agreed text encoding. The symbols $r_1,r_2,r_3$ are unspecified, so there are no numerical message blocks to evaluate further.

Put $E=F(a,b)$. The tower law for field extensions shows that both $m$ and $n$ divide $[E:F]$. Since $m,n$ are coprime, their product divides $[E:F]$, so $[E:F]\ge mn$.

The minimal polynomial of $b$ over $F$ has degree $n$ and is still an annihilating polynomial over $F(a)$. The minimal polynomial over $F(a)$ divides it, soThe upper and lower bounds agree:Coprimality is the step forcing the lower bound to equal the product; without it the two generated fields can overlap substantially.

Write the group action as $(\omega ,g)\mapsto \omega g$. A right group action satisfies $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. Its orbit and stabilizer areThe identity belongs to $H$. If $g,h\in H$, then $\omega gh=(\omega g)h=\omega$, so $gh\in H$. Also $\omega g=\omega$ implies $\omega =(\omega g)g^{-1}=\omega g^{-1}$, so $g^{-1}\in H$. Thus $H$ is a subgroup.

For the orbit-stabilizer theorem, the map $G\to \mathrm{Orb}(\omega )$, $g\mapsto \omega g$, is onto, andIts fibers are precisely the right cosets $Hh$, each containing $|H|$ elements. Counting these disjoint fibers givesFor the symmetric group $S_3$, the six subgroups areThe group action $H\mapsto g^{-1}Hg$ has four orbits: each of $\{1\}$, $A_3$, and $S_3$ is fixed, while the three order-two subgroups form one orbit. Here an orbit is a collection of subgroups, rather than a collection of individual permutations.