For an oriented piecewise smooth surface $S$ with boundary $C=\partial S$ and a vector field $\mathbf{A}$ with continuous first derivatives near $S$, the Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector, $dS$ the positive area element, and $d\mathbf{r}$ the oriented tangent line element. The traversal of $C$ follows the right-hand rule relative to the surface orientation.

The comparison test for series says that $0\le v_n\le C{w}_n$ eventually and convergence of $\sum w_n$ imply convergence of $\sum v_n$. Conversely, an eventual positive lower comparison with a divergent positive series proves divergence. The ratio test uses $L={\lim }_{n\to \infty }{v}_{n+1}/v_n$: $L<1$ proves convergence, $L>1$ proves divergence, and $L=1$ is inconclusive. Removing finitely many initial zero terms does not affect these conclusions.

Use cylindrical coordinates with $dV=rdrd\theta dz$, $0\le \theta <2\pi$, $0\le z\le h$ and $0\le r\le a(1-z/h)$. Set $s=1-z/h$. Integrating first over radius gives the mass per unit heightThereforeFor the first moment about the base, use $z=h(1-s)$:Dividing by the total mass givesThe center of mass of a cone with a power-law radial density gives an independent decomposition: the uniform-density component has centroid $h/4$, while the component proportional to $r$ has centroid $h/5$ and half as much mass. Their mass-weighted mean is $7h/30$. Density increasing toward the wider outer parts consequently moves the center below the uniform cone's $h/4$.

In components, the scalar triple product isIt is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$. A cyclic permutation of the rows has positive sign, so $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$. Equivalently the Levi-Civita symbol expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by that cyclic relabeling.

Expand $u=3x^2-2xy+3y^2$. The Lagrange multiplier equations for $u-\lambda v$ are $(6-2\lambda )x-2y=0$ and $-2x+6y+\lambda =0$, together with the constraint $y=x^2-1/4$. The first two imply $y=(3-\lambda )x$ and $(6\lambda -16)x=\lambda$. The value $\lambda =8/3$ would make the latter equation inconsistent, so division is allowed:Insert these into $x^2-y-1/4=0$ and multiply by $(6\lambda -16{)}^2$. Simplification gives $2(3{\lambda }^3-21{\lambda }^2+48\lambda -32)=0$, henceThe gradient of the constraint $(2x,-1)$ is never zero, so the Lagrange multiplier condition is necessary at every constrained extremum. To establish a global minimum, parametrize the whole parabola by $y=x^2-1/4$. The restricted function becomesIt is coercive: $q(x)\to \infty$ as $|x|\to \infty$, so it attains a global minimum. Moreover $q^{\prime \prime }(x)=36(x-1/6{)}^2+2>0$, making it strictly convex and ensuring that this minimum is unique. This proves more than the permitted assumption of one real multiplier root.

The level sets of $u$ are nested ellipses centered at the origin. In the rotated coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, they have equation $2p^2+4q^2=\text{constant}$, with the longer axis along $x=y$. The smallest one meeting the parabola is tangent to it at the unique minimizing point, approximately $(x,y)=(-0.12654,-0.23399)$, where $u\simeq 0.15307$ and $\lambda \simeq 1.15085$.