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For an oriented surface $S^{\prime }$ with compatible boundary direction, the Stokes theorem statesLet $H$ denote the removed patch and $S^{\prime }=S\setminus H$ the remaining cylinder. Keep the outward orientation inherited from the closed cylinder. Everywhere on the removed curved patch, the normal curl component is $-\alpha R$, so its flux is $-\alpha RA$, where $A$ is its actual curved area. Part (d) then givesConsequently the circulation around a hole in an oriented surface isThe positive direction here is induced by the remaining outward-oriented surface; locally it runs clockwise around the hole when viewed from outside. If the edge is instead traversed positively around the removed patch, the direction reverses and the answer is $-\alpha RA$. The question does not specify an edge direction, so the sign must be accompanied by this convention. The result depends only on the removed curved area, not on its shape. If the patch has several boundary components, use the sum of their induced line integrals. Since $[\alpha ]=1/(\text{length}\cdot \text{time})$, the result has the units of a velocity line integral, length squared per time.

Put $L=5\pi /2$. The curve tangent is ${\mathbf{r}}^{\prime }=(-b\sin \theta ,b\cos \theta ,c)$. For the first line integral, the scalar triple product identity gives $(\mathbf{a}\times \mathbf{r})\cdot {\mathbf{r}}^{\prime }=\mathbf{a}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })$. Its cross product isThe needed definite integrals areand ${\int }_0^L(-\theta \sin \theta -\cos \theta )d\theta =[\theta \cos \theta -2\sin \theta {]}_0^L=-2$. HenceFor the second line integral, substitute the curve into ${\mathbf{v}}_2\cdot {\mathbf{r}}^{\prime }$. The cubic terms combine to $b^3\cos \theta$, leavingThe fundamental theorem of calculus therefore givesBoth calculations retain the full path from $\theta =0$ to $5\pi /2$, rather than inadvertently closing the helix after one turn.

On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

The modified upper surface has exactly the same ellipse at $z=0$ and the same boundary orientation. The field retains the global polynomial vector potential from part (a). By the Stokes theorem, both surface fluxes equal the same boundary line integral; the cap's height and interior shape do not enter. HenceThis is an instance of the fact that solenoidal flux depends on the oriented boundary, with the domain condition automatically satisfied by the smooth polynomial field on all of ${\mathbb{R}}^3$. When $a,b$ are chosen positive, the formula simplifies to $\pi a^{3}b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.

The fundamental theorem for line integrals evaluates this conservative vector field integral by the change in potential. At the initial point $\varphi =0$. At the final point $z=0$, so the exponential factor is one, andThe detailed shape of the path does not affect this line integral.

There is a global-domain issue in the printed question. Write $D=a^2{x}^2+b^2{y}^2>0$ away from the origin. Direct partial differentiation givesThus the differential one-form is a closed differential form for every real $a,b\ne 0$. The Poincaré lemma makes it locally an exact differential form on small neighborhoods avoiding the origin. There is no additional condition such as $a^2=b^2$.

However, the stated domain is the whole punctured plane, which is not simply connected. To test global exactness, use the closed ellipse $x=\cos t/a$, $y=\sin t/b$, $0\le t\le 2\pi$. Here $D=1$ andThe line integral of a globally exact differential form around any closed curve would be zero. ConsequentlyThe distinction is precisely that of the elliptically scaled angular one-form. The remaining parts can be answered consistently using a local angle branch, as follows.

Evaluate the line integral along the three coordinate segments, treating coordinates not changing along a segment as constants. On the first segment, $y=z=0$, so $F_x=0$ and the contribution is zero. On the second segment,On the final segment,The fundamental theorem for line integrals therefore givesTaking its gradient verifies each component of the field. The constant is $C=\Phi (0,0,0)$; the field specifies only potential differences because the gradient of every constant is zero.

For $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is ${\mathbf{r}}^{\prime }(t)=(1,1,2t)$. Substitute into the field:Its scalar product with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized line integral isAs a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.

Starting at $(1,0)$, traverse the right, top, left and bottom sides in that order. For $\mathbf{F}$, the differential expression is $-ydx+xdy$. The four line integral contributions are respectively $1,1,0,0$, so the total is two.

For $\mathbf{G}$, the expression is $2x{y}^{2}dx+2y{x}^{2}dy$. The right side contributes ${\int }_0^{1}2ydy=1$, and the top side contributes ${\int }_1^{0}2xdx=-1$; the other sides contribute zero. Therefore

The field $\mathbf{F}$ is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=1-(-1)=2$.

For $\mathbf{G}$, integrating ${\Phi }_x=2x{y}^2$ gives $\Phi =x^2{y}^2+h(y)$. Matching ${\Phi }_y=2y{x}^2$ then gives $h^{\prime }=0$. ThusThis also explains why both closed line integrals of $\mathbf{G}$ vanish.

The arc length factor is the norm of the tangent:Hence $ds=10dt$ and $\Phi =xy=\cos 6t\sin 6t=\frac{1}{2}\sin 12t$. The scalar line integral is thereforeThis uses scalar arc length, rather than the oriented displacement used in the vector line integrals of part (a).

A line integral uses $\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)$, including the tangent vector. With $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, the three components sum toConsequently

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 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.

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).

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.

The original PDF has angle $\theta (t)=2\pi t^n$, with $n$ an exponent on $t$. It is not $2\pi nt$: each permitted value of $n$ changes the speed but traverses the same circle once counterclockwise. Along the path, $r^2=2$ andThe radial contribution to the line integral vanishes, since $(\cos \theta ,\sin \theta ,1)\cdot (-\sin \theta ,\cos \theta ,0)=0$. The rotational contribution givesThereforeThe answer is independent of the monotone parametrization, as an oriented line integral should be.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

Put $q=2t-t^2$. The second coordinate is exactly $q^2$, so this is the same parabola as in part (i), with a different increasing parametrization. Since $q$ runs from zero to one and $dq=(2-2t)dt$, the line integral isEquivalently, substitute the original parameter to get ${\int }_0^{1}5(2t-t^2{)}^3(2-2t)dt$. This demonstrates that a line integral depends on the oriented path, not on its speed of traversal.

With $\mathbf{r}(t)=(t,t^2{)}^T$, the tangent vector is $(1,2t{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,2t^2{)}^T$. The line integral is