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For the parametric curve $\mathbf{r}(t)=(t,t^2,t^3)$, $0\le t\le 1$, the vector field and the tangent vector areConsequently the line integral is

Take counterclockwise traversal as viewed from $z>0$. In the plane $z=0$, the work is the planar line integral of $P=2x-y$ and $Q=x+y$. Green's theorem givesThe $z$ component contributes nothing because $dz=0$. The question specifies one circuit without specifying its direction: clockwise traversal gives $\boxed{-18\pi }$.

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.

Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

A parametrization of the straight line segment is $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then $\mathbf{F}(\mathbf{r}(t))=(t,t^2,t)$ and ${\mathbf{r}}^{\prime }(t)=(1,1,1)$. By the line integral formula,

For the curved path $r(t)=(t,t^2,t^3)$, $r^{\prime }(t)=(1,2t,3t^2)$. The vector field along this parametric curve isIts dot product with the tangent is $(t^2+2t)e^t+13t^{12}$. Therefore the direct line integral is

Parametrize the straight line by $r(t)=(t,t,t)$, $0\le t\le 1$, with $r^{\prime }(t)=(1,1,1)$. Substituting into the vector field and taking the dot product givesHence the direct line integral is

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$. Its tangent is $(-1,1,0)$ and $\mathbf{F}=(-t,1-t,t-1)$, so $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=1$. Hence its line integral isFor the other line integral, $|\mathbf{r}{|}^2=(1-t{)}^2+t^2$ and $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2t-1$. ThusThis also equals the change in $\Phi$, by the fundamental theorem for line integrals.

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

Parametrize the line by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Its tangent vector is $(1,1,1)$. On this line the three force components sum to $\mu t+(1-\lambda )t+(\lambda -1-\mu )t=0$. Therefore the line integral is $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$ for every parameter choice.

Orient the triangle by the path $A\to B\to C\to A$. The cross product $(B-A)\times (C-A)=(0,1,0)$ gives the positive normal vector. In the plane $y=1$, the triangular region is $x,z\ge 0$, $x+z\le 1$. The surface integral in Stokes theorem is thereforeFor the boundary line integral, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\le t\le 1$. Its contribution isThus both oriented integrals equal $1/15$. Reversing the orientation would reverse both signs.

On the first segment only $z$ varies and $F_z=x^{2}y=0$. On the second only $y$ varies and $F_y=x^{2}z=0$. On the last segment $y=z=\pi$, so the line integral isThe agreement is the expected path independence of a conservative vector field.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le \pi$. Then ${\mathbf{F}}_3\cdot d\mathbf{r}=(4t^3+\sin t)dt$, so direct evaluation of the line integral givesThe fundamental theorem for line integrals supplies an independent check: this is ${\varphi }_3(\pi ,\pi ,\pi )-{\varphi }_3(0,0,0)$.

Parametrize the straight line by $\mathbf{x}(t)=(t,t,t)$, $0\le t\le 1$. The line integral isThe change in scalar potential is $\psi (1,1,1)-\psi (0,0,0)=-3$, so the answer is $-\Delta \psi$, as required by the fundamental theorem for line integrals for $\mathbf{F}=-\nabla \psi$.

The parametric curve runs from $t=0$ to $t=1$, with ${\mathbf{r}}^{\prime }=(2,3t^2,6t)$. Substitution into the two line integrals givesHenceThe first is unchanged because the field is conservative. The two different answers for the second exhibit its path dependence, consistent with its nonzero curl.

On the first segment, $x=z=0$, so both line integrals vanish. On the second, $y=1,z=0$, giving contributions ${\int }_0^{2}3x^{2}dx=8$ and ${\int }_0^{2}2xdx=4$. On the last, $x=2,y=1$, the contributions are ${\int }_0^3-2zdz=-9$ and ${\int }_0^3-6z^{2}dz=-54$. ThereforeThe first also equals $\varphi (B)-\varphi (A)=8-9=-1$, as the fundamental theorem for line integrals predicts.

Parametrize the directed segment by $\mathbf{r}(t)=(t,t,0)$, $0\le t\le 1$. Its tangent is $(1,1,0)$, while the restricted vector field is $(t^3,t^3,0)$. The line integral is thereforeAlternatively, $\mathbf{G}=\nabla (x^2{y}^2/2)$, so the fundamental theorem for line integrals gives the same endpoint difference.

Put $R=\sqrt{8}$ and parametrize the circle counterclockwise, as viewed from positive $z$, by $\mathbf{r}(\varphi )=(R\cos \varphi ,R\sin \varphi ,3)$ for $0\le \varphi \le 2\pi$. The restricted field isThe PDF does not specify the differential in this line integral. An oriented vector line integral weighted by a scalar field uses $d\mathbf{r}=(-R\sin \varphi ,R\cos \varphi ,0)d\varphi$. Its two nonzero components areAll other terms vanish by sine-cosine symmetry. Thus this interpretation gives

If the intended convention is the scalar arc length integral, then $ds=Rd\varphi$ andThat scalar integral is independent of traversal direction. Giving both values retains the actual distinction left unspecified by the wording.

For $\mathbf{r}(t)=(t,t,2t^3)$, the conservative field is $(6t^7,4t^7,t^5)$ and the tangent is $(1,1,6t^2)$. ThereforeThe two different paths give the same value because their line integrals depend only on the potential difference ${\varphi }_1(B)-{\varphi }_1(A)=2$. There is no scalar potential for ${\mathbf{F}}_2$ to which the conditional request could be applied.