Take the outward orientation on the upper hemisphere; its boundary circle is then counterclockwise viewed from above. Parametrize that circle by $r(\varphi )=(r\cos \varphi ,r\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Then $G\cdot dr=r^2(3{\cos }^2\varphi -4{\sin }^2\varphi )d\varphi$, so Stokes theorem givesEquivalently $\nabla \times G=(0,0,-1)$, and the upward projected area of the hemisphere is $\pi r^2$. The opposite orientation gives the negative of this answer.

The divergence is $\nabla \cdot F=2xyz+2xyz+2xyz=6xyz$. With outward orientation, the divergence theorem converts the surface flux into a volume integral:Directly, the three faces lying in coordinate planes contribute zero, and each opposite face contributes $1/4$, confirming the sign and normalization.

For a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, oriented by the outward unit normal vector, and a continuously differentiable vector field on a neighborhood of $\overline{V}$, the divergence theorem statesFor an oriented piecewise smooth surface $S$ with piecewise smooth boundary $C=\partial S$, and a continuously differentiable vector field on a neighborhood of $S$, Stokes theorem statesThe boundary orientation follows the right-hand rule relative to the chosen normal vector. Reversing that orientation reverses both sides of the latter identity.

The characteristic polynomial is $(5-\lambda )(2-\lambda )-4={\lambda }^2-7\lambda +6=(\lambda -1)(\lambda -6)$. For $\lambda =1$, the eigenvector equation gives $y=2x$; for $\lambda =6$, it gives $x=-2y$. The normalized eigenvectors can therefore be chosen asBoth have unit Euclidean norm, and their dot product is $(-2+2)/5=0$, explicitly verifying orthogonality. This also follows from the spectral theorem for a real symmetric matrix with distinct eigenvalues.

The unique symmetric and antisymmetric parts of a matrix are $S=(M+M^T)/2$ and $A=(M-M^T)/2$, givingTheir matrix products with their transposes are $S^{T}S=(a+b{)}^{2}I/4$ and $A^{T}A=(a-b{)}^{2}I/4$. Each is an orthogonal matrix precisely when the corresponding factor is one. Thus $a+b=\pm 2$ and $a-b=\pm 2$, with independent sign choices, and the simultaneous possibilities are

Write $c=\cos \theta$, $s=\sin \theta$. Since $\theta$ is constant, the multivariable chain rule gives ${\partial }_x=c{\partial }_u-s{\partial }_v$ and ${\partial }_y=s{\partial }_u+c{\partial }_v$. Applying these operators twice to a twice continuously differentiable function givesThe mixed terms cancel and $c^2+s^2=1$, soThis is rotational invariance of the Laplacian: an orthogonal change of coordinates preserves the trace of the Hessian matrix. A nonconstant angle would introduce additional derivatives and is not covered by this constant rotation calculation.

Away from the origin, the chain rule gives $r_x=x/r$ and $r_y=y/r$. Differentiating once more yields the two-dimensional case of the Hessian of the Euclidean norm:These partial derivatives are smooth wherever $r>0$, so Clairaut's theorem gives $\boxed{r_{yx}=r_{xy}=-xy/r^3}$. None of these formulas supplies a derivative at the origin: even the restriction $r(x,0)=|x|$ has no two-sided derivative there.

The displacement vectors are $B-A=(-1,2,-2)$ and $C-A=(1,0,1)$. Their cross product is $(2,-1,-2)$, with Euclidean norm three. Thus either sign ofis a unit normal vector to the second plane, whose equation is $2x-y-2z=3$. A direction vector for the intersection is the cross product of the two normal vectors:It is orthogonal to each normal vector. Since both planes contain $A$, the whole intersection line is $r=(2,-1,1)+t(6,2,5)$, $t\in \mathbb{R}$.

With normal vector $n=(-1,-2,2)$, a plane through $A=(2,-1,1)$ has equation $n\cdot (r-A)=0$. As $n\cdot A=2$, its plane equation isThe distance from a point to a plane is the absolute signed offset divided by the Euclidean norm of its normal vector. Hence the Euclidean distance from the origin is $\boxed{2/3}$, since $|n|=3$. The closest point is $2n/9=(-2,-4,4)/9$, whose Euclidean distance from the origin is $2/3$ and which satisfies the plane equation.

Put $u=\cos x$, $du=-\sin xdx$, and use ${\sin }^{2}x=1-u^2$. ThenThereforeThis holds on intervals where $\cos x\ne 0$.