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The first field is ${\mathbf{F}}_1=(z-y,x,-x)$, so direct curl calculation givesIt is not conservative because its curl is nonzero. The second is a radial gradient field:The scalar expression is a smooth function and single-valued on all of space, so ${\mathbf{F}}_2$ is a conservative vector field. Its potential of a conservative vector field is $\Phi =-(1+r^2{)}^{-1/2}$. Parts (a) and (b) check path dependence directly.

Use the real Fourier series convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$. The function is an even function, so $b_n=0$. For noninteger real $\mu$,For $n\ge 1$, the product-to-sum identity givesThus the Fourier series of a noninteger-frequency cosine isIts Fourier coefficients are $O(n^{-2})$, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting $\mu =1/3$ givesEvaluate at $x=0$, where $f(0)=1$, and multiply by $\pi /\sqrt{3}$:Comparing with the required form gives $\boxed{A=3,B=3}$.

Differentiate the three components with respect to their matching Cartesian coordinates:The $3x{z}^2$ term contributes nothing to the $y$ derivative because it is independent of $y$.

The dot product is $\mathbf{a}\cdot \mathbf{b}=-1+2+3=4$, while the Euclidean norms are $\sqrt{3}$ and $\sqrt{14}$. Thus the angle in the stated range isThe positive cosine shows that it is an acute angle.

Isolate the square root: $\sqrt{2x^2+17}=3-x$. Its right side must be nonnegative, so retain the condition $x\le 3$ before squaring. Squaring then givesBoth candidates pass the original equation: $\sqrt{25}-2=3$ and $\sqrt{49}-4=3$. ThusThe substitution check excludes the possibility of extraneous roots introduced by squaring.