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The solid angle is the intercepted surface area on a sphere of radius $R$, divided by $R^2$. Take the cone axis as the polar axis of spherical coordinates, and call its half-opening angle $\theta$. With polar integration variable $\vartheta$,For small $\theta$ measured in radians, the Taylor series of the cosine givesThis is the solid angle of a circular cone, measured in steradians.

The supplied angular diameter is twice the half-opening angle, soHence either apparent disc subtendsThe fraction is about $0.0005409\%$ of the entire celestial sphere. Using the angular diameter directly in $\pi {\theta }^2$ would give four times the correct small-angle answer.

A normal mode is a motion of a linear oscillatory system in which every coordinate has the same angular frequency and a fixed relative amplitude and phase. For a conservative small-displacement system it has the formwhere $q$ is the mode shape, $K$ the stiffness matrix, and $M$ the mass matrix. Superposing normal modes gives the general small-oscillation motion when their eigenvectors form a complete basis.

Two common examples are standing transverse oscillations of a string fixed at both ends, and oscillations of coupled masses and springs. The string's mode shapes are sinusoidal along its length; the coupled system's mode shapes are fixed patterns of particle displacements. The following parts determine those patterns for the three-mass system in the source figure.

The successive derivatives are $f^{(n)}(x)=(-6{)}^n{e}^{-6x}$, so $f^{(n)}(-4)=(-6{)}^n{e}^{24}$. Substitution in the Taylor series formula about $a=-4$ givesThe ratio test shows convergence for every finite real or complex $x$: the absolute ratio of successive terms is $6|x+4|/(n+1)\to 0$. This is also the usual exponential function series applied to $e^{24}{e}^{-6(x+4)}$, proving that the series equals the stated function everywhere.

Assume small planar oscillations about the vertical equilibrium, with the connecting spring relaxed in that equilibrium. Let $x_j=l{\theta }_j$ be the first-order horizontal displacements of the two simple pendulums. The gravitational restoring force on bob $j$ is $-(mg/l)x_j$; the spring extension to first order is $x_2-x_1$. Thus the linearized equations of motion areAdding and subtracting them diagonalizes the system. For $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$,The normal modes and angular frequencies are thereforeThe in-phase mode does not stretch the spring, while the opposite-phase mode adds a restoring force of magnitude $2k$ times each bob's displacement. These are the small-oscillation frequencies, not exact amplitude-independent frequencies for arbitrary pendulum angles.

For an oriented piecewise smooth surface $S$, a continuously differentiable vector field $F$ on a neighborhood of it, and its compatibly oriented boundary $C=\partial S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along $C$. The right-hand rule relates the boundary direction to $n$.

The supplied field has curl $\nabla \times F=(2z,2x,2y)$. Parametrize the plane by $r(x,y)=(x,y,1-x-y)$ over the triangle $D=\{x,y\ge 0:x+y\le 1\}$. Its upward oriented surface area element is $(1,1,1)dxdy$, givingFor the boundary line integral, take the positively oriented vertices $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Their projections run counterclockwise in the horizontal plane. With $0\le t\le 1$, the three edges areOn each edge, direct substitution gives $F(r(t))\cdot r^{\prime }(t)=t^2+2t-1$. HenceThe two independent computations agree, including the orientation sign.

A conservative vector field can be written as $F=\nabla \Phi$ for a single-valued scalar potential $\Phi$. Its line integral between two points depends only on the endpoints, and its integral around every closed curve is zero. A necessary local test for a smooth planar field $F=(P,Q)$ is $P_y=Q_x$. On a simply connected domain such as the whole plane, this is also sufficient.

HereAn explicit scalar potential is even more decisive:Indeed ${\Phi }_x=2x{e}^{xy}+x^{2}y{e}^{xy}$ and ${\Phi }_y=x^3{e}^{xy}+2y$. Thus the field is conservative on ${\mathbb{R}}^2$.

An exact differential is a differential one-form equal to $d\Phi$ on its domain; an inexact differential is not the differential of a single-valued potential there. For example, $d(xy)=ydx+xdy$ is exact. For the supplied form $\eta =42ydx+xdy$, the cross derivatives areThey disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise line integral around the unit square is $1-42=-41\ne 0$. ThereforeThe connection is $F\cdot dr=d\Phi$: a vector field is conservative precisely when its work differential one-form is exact. Locally vanishing curl alone need not guarantee global exactness on a domain with holes.