From $X=s^2-4$ and $Y=4s$, elimination of $s$ givesThis makes the real part a quadratic function of the imaginary part. There is no sign ambiguity because $Y$ determines $s=Y/4$ uniquely. The calculation illustrates power maps of a reciprocal-coordinate hyperbola.

The quadratic formula gives $t=-1\pm \sqrt{2}$. The tangent function is positive in the first quadrant, so the positive polynomial root is selected:

The double-angle formula for the tangent function gives $1=2t/(1-t^2)$. Here $0<t<1$, because $0<\pi /8<\pi /4$, so the denominator is nonzero. Rearranging yields the quadratic equation

Take a piecewise linear path along three coordinate directions, passing through the origin and then $(0,1,0)$:On the first segment $G_x=y^2+z^2=0$, so its line integral is zero. On the second $G_y=x^2+z^2=0$, so its contribution is also zero. On the last segment $G_z=x^2+y^2=1$ and $z$ increases from zero to one. ThereforeThis gives the requested concrete path dependence with the same ordered endpoints.

For the second vector field, the curl isIt is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable scalar potential would have zero curl of its gradient, so $\mathbf{G}$ is not conservative. The fact that its curl vanishes on the line $x=y=z$ does not make it vanish throughout the domain.

The three components of the curl cancel pairwise:The domain is all of ${\mathbb{R}}^3$, which is simply connected, so this is consistent with a conservative vector field. To construct its scalar potential, integrate ${\Phi }_x=y+z$ to obtain $\Phi =x(y+z)+h(y,z)$. Matching ${\Phi }_y=x+z$ gives $h_y=z$, hence $h=yz+k(z)$. Finally ${\Phi }_z=x+y$ forces $k^{\prime }(z)=0$. ConsequentlyEvery scalar potential differs by a constant on this connected domain. The endpoint values differ by one, agreeing with the line integral in part (a).

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

Parametrize the segment in its stated direction by $\mathbf{x}(t)=(1-t,t,t)$, $0\le t\le 1$. Then $d\mathbf{x}=(-1,1,1)dt$, and substitution into the two vector fields givesTaking each dot product with $d\mathbf{x}$ evaluates the line integrals:

The new cone can be parametrized byThe outward vector area element is $r(\frac{b}{a}\cos \theta ,1,\frac{b}{a}\sin \theta )drd\theta$. This is ${\mathbf{R}}_{\theta }^{\prime }\times {\mathbf{R}}_r^{\prime }drd\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its dot product with $\mathbf{F}({\mathbf{R}}^{\prime })$ is again $brdrd\theta$. HenceThis equality also follows because a rigid rotation takes the first cone into the second and leaves the radial vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ invariant.

On the sphere of radius $a$, the outward unit normal is $\mathbf{n}=(x,y,z)/a$. Thus $\mathbf{F}\cdot \mathbf{n}=a$ everywhere. Multiplying by the surface area $4\pi a^2$ givesThe same surface integral follows in spherical coordinates from $dS=a^2\sin \vartheta d\vartheta d\phi$.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.