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The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

Expanding the determinant along the first row givesThe matrix trace is the sum of the diagonal entries, $\mathrm{tr}A=1+4+1=6$. Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus $A$ is singular; the determinant alone does not specify the two other eigenvalues.

For the partial fraction decomposition, write $A/(x-4)+B/(x+9)$. Clearing denominators gives $13(x+1)=A(x+9)+B(x-4)$. Substitution at $x=4$ and $x=-9$ yields $A=5$ and $B=8$, hence

The original line meets the horizontal axis at $(-3/2,0)$. A perpendicular line has slope $-1/2$, the negative reciprocal of $2$, soSubstituting $x=-3/2$ verifies the required common point.

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

Write the rational expression as $(x^2+4{)}^{-1}$ and apply the chain rule:The denominator is positive for every real $x$, so the derivative is defined throughout the real line.

The product rule for divergence, $\nabla \cdot \mathbf{r}=3$, and the result in (a) give, for $r>0$,This is a pointwise identity away from the origin, where the radial unit vector is undefined.

For $r>0$, use the gradient of a radial function with $g(r)=1/r$:Equivalently each component is $-x_j/r^3$ by differentiating $r^2=x^2+y^2+z^2$. The origin is excluded from the function's domain.

For the divergence in Cartesian coordinates, differentiate each component with respect to its own coordinate:The third component contributes zero because it is independent of $z$.

First take $F_y=x{e}^{\sin (xy)}\cos (xy)$, then differentiate with respect to $x$. The product rule and chain rule giveThe function is smooth, so the two mixed partial derivatives agree.

Project the radius vector onto the $x$-$y$ plane. The polar angle $\theta =\pi /6$ lies between the positive $z$ axis and $OP$. The azimuth $\varphi =\pi /3$ lies in the $x$-$y$ plane between the positive $x$ axis and the projection of $OP$, measured toward positive $y$.

Use the convention that $\theta$ is the polar angle from positive $z$, and $\varphi$ the azimuth from positive $x$. The radius isThen $\cos \theta =z/r=\sqrt{3}/2$, while $\tan \varphi =y/x=\sqrt{3}$ and both planar coordinates are positive. Thus the spherical coordinates are

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.

For $m>-1$, integrate the base power to obtain $J(m)={\int }_0^1{x}^{m}dx=1/(m+1)$. Since ${\partial }_m^n{x}^m=x^m(\log x{)}^n$, differentiating $n$ times givesHere $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x{|}^n$, whose integral becomes ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$ under $x=e^{-t}$ and is finite. The dominated convergence theorem establishes the needed differentiation under the integral sign and the logarithmic moments of a power on the unit interval.

Under the regularity and integrability conditions for the Leibniz integral rule,The first two terms account for the moving endpoints; the last accounts for the integrand's own parameter dependence.

For a real number $x$, the hyperbolic functions satisfy $\cosh x>0$ and $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$. Hence $\cosh x=1/\sqrt{1-u^2}$, and multiplying it by $\tanh x=u$ givesThe sign of the numerator preserves the sign of $\sinh x$.

The defining exponential function formulas for the hyperbolic functions give

First calculate the curl:By Stokes theorem, use the positively oriented equator $\mathbf{r}(\varphi )=(a\cos \varphi ,a\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Its tangent is $a(-\sin \varphi ,\cos \varphi ,0)$, while the restricted field is $(0,3a^2\cos \varphi \sin \varphi -a\cos \varphi ,0)$. ThusThe first term integrates to zero over a full period and the second uses ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$.

One can also calculate directly on the curved surface. With spherical coordinates $0\le \theta \le \pi /2$, $0\le \varphi \le 2\pi$,Substitution gives the scalar integrand $a^3{\sin }^2\theta \cos \theta \sin \varphi -a^2\sin \theta \cos \theta$. Its azimuthal sine term vanishes, and the rest gives $-2\pi a^2{\int }_0^{\pi /2}\sin \theta \cos \theta d\theta =-\pi a^2$, confirming the orientation and the same result.

For a smooth oriented surface $S$ with positively oriented boundary $C=\partial S$, Stokes theorem statesHere $d\mathbf{S}=\mathbf{n}dA$ uses the chosen unit normal vector, while $d\mathbf{r}$ is tangent to the directed boundary. The boundary orientation follows the right-hand rule: curling the fingers along $C$ makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the surface, with the usual piecewise-smooth allowance for its boundary.

For the upper hemisphere with outward normal, the induced equatorial direction is counterclockwise as viewed from positive $z$.

The upper branch leaves the origin horizontally and curves upward, growing like $x^{5/2}$. The lower branch shares the horizontal tangent, rises to its small maximum at $x=16/25$, then crosses the axis at $x=1$ and falls toward $-\infty$. Its concavity changes at $x=64/225$, before the maximum. A separate zoom makes those nearby lower-branch features visible.