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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.

Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

Apply the multivariable chain rule to $f(x,y)=g(r(x,y),\varphi (x,y))$. Using the coordinate partial derivatives above givesEquivalently, ${\partial }_x=\cos \varphi {\partial }_r-(\sin \varphi /r){\partial }_{\varphi }$ and ${\partial }_y=\sin \varphi {\partial }_r+(\cos \varphi /r){\partial }_{\varphi }$. Their coefficients vary with position and must also be differentiated when applying them twice.

Differentiate the angular relation locally, or use $d\varphi =(xdy-ydx)/(x^2+y^2)$. Holding each other coordinate fixed givesThese are local partial derivatives on a consistent angular branch, independent of where its branch cut is placed.

For $r>0$, differentiate $r=\sqrt{x^2+y^2}$ while keeping the other coordinate fixed in the Cartesian coordinate system:The origin is excluded from this polar coordinates chart.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.

Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

The expected value follows by integration by parts:The same result follows from symmetry of the probability density function about $\pi /2$.

Normalization of the probability density function requires $1=\alpha {\int }_0^{\pi }\sin xdx=2\alpha$. Thus

The plane contains $(1,0,1)$ and has normal vector $(0,1,2)$, so its equation is $y+2z=2$. In the equation $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal, $|d|$ is the perpendicular distance from the origin. Consequently