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The Stokes theorem states that for an oriented smooth surface $S$ with compatible oriented boundary $C$ and a continuously differentiable vector field,Here $\nabla \times F=(-2,-2,-2)$, whereas a normal vector to the specified plane is $(1,0,-1)$. Their dot product is zero. Thus every closed curve in that plane has zero circulation by the Stokes theorem.

The source says “any curve” and displays an ordinary line integral, without specifying that it is closed or giving endpoints. An open curve need not give zero. On the plane, put $z=x-1$ and $dz=dx$; the pulled-back differential one-form becomesTherefore the complete endpoint-dependent answer iswhich specializes to zero for a closed curve. For example, the path from $(1,0,0)$ to $(2,0,1)$ lies in the plane and has integral $1$. This illustrates a conservative restriction of a vector field to a plane despite nonzero ambient curl.

Holding $y$ fixed gives $u_x=v_x=1$, whereas holding $x$ fixed gives $u_y=1$ and $v_y=-1$. The chain rule therefore gives $w_x=w_u+w_v$ and $w_y=w_u-w_v$. Multiplication yieldsThe subscripts specify the independent coordinate held fixed; they are necessary because changing coordinates changes which direction a partial derivative follows.

A standing wave has a fixed spatial amplitude profile and oscillates in time, with stationary nodes of a standing wave. It arises from equal-amplitude waves of the same frequency travelling in opposite directions. For a uniform string with linear mass density $\mu$, the wave speed is $c=\sqrt{T/\mu }$ and a separated mode has $Y^{\prime \prime }+k^{2}Y=0$.

The fixed boundary requires $Y(0)=0$, while the free boundary requires $T{Y}^{\prime }(L)=0$. Thus $Y=A\sin kx$ and $\cos kL=0$, givingThe fixed end is a node of a standing wave and the free end an antinode. Arbitrary small motions are superpositions of these normal modes; the tension alone determines their frequencies only when $\mu$ is also specified.

Let $G(s)={\int }_0^s{e}^{t^2}dt$. The fundamental theorem of calculus gives $G^{\prime }(s)=e^{s^2}$. Applying the chain rule to $F(x,y)=G(x{y}^2)$ givesNo elementary antiderivative of $e^{t^2}$ is needed: differentiation of the variable upper limit supplies the integrand directly.

Stokes theorem states that ${\oint }_{C}A\cdot dr={\iint }_S(\nabla \times A)\cdot ndS$, with consistent boundary orientation. Here $\nabla \times A=e_z$, so the circulation around the positively oriented planar curve equals its area. Substituting the given parametrization givesThe curve closes at the origin, passes along the right side to $(0,1)$ and returns along the left side, so it is anticlockwise. ConsequentlyEquivalently its two sides are $x=\pm 2y\sqrt{1-y^2}$, $0\le y\le 1$, whose horizontal width integrates to the same result.

Set $x=au$, $y=bv$, $z=cw$. For nonzero $a,b,c$, the absolute Jacobian determinant is $|abc|$, and the transformed interior is the ball $u^2+v^2+w^2\le 4$. Its radius is two, soIf $a,b,c$ are taken as positive semiaxis parameters, this is $32\pi abc/3$. The actual semiaxes are $2|a|,2|b|,2|c|$; omitting the factor two in the transformed radius would make the volume eight times too small.