The field is smooth on the simply connected domain ${\mathbb{R}}^3$. Its curl is $\nabla \times \mathbf{F}=(2\lambda -2,2\mu ,0)$, so a necessary and sufficient condition for a conservative vector field isAt those values it is explicitly the gradient of $\Phi =xy-x^2/2-yz+z^2/2$. Equality of the two previous path integrals would impose only $\lambda -\mu =1$, which is weaker: coinciding path integrals need not imply a conservative field. For example $\lambda =2,\mu =1$ makes those two integrals equal but leaves nonzero curl.

On the specified curve $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is $(1,1,2t)$ andTheir dot product is $t+(\lambda -\mu -4)t^2+2t^3$. Hence

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.

Use the slope substitution for equations involving x y'-y: put $v=y/x$ on the positive interval containing the initial point. Then $x{y}^{\prime }-y=x^2{v}^{\prime }$, so $x^2{v}^{\prime }=\cot v$. Near $v(1)=\pi /4$, separating variables gives $\tan vdv=dx/x^2$. Integration yields $\ln |\cos v|=1/x+C$. The initial value fixes the sign and constant, givingChoose the inverse cosine branch that agrees with $v(1)=\pi /4$. The solution isOn this interval the argument lies strictly between zero and one and the original cotangent is defined. At the finite lower endpoint, $v\to 0$ and the cotangent becomes singular, so that endpoint is excluded from the maximal interval of the initial value problem. The slope substitution here is valid without classifying the original equation as homogeneous in the usual $y^{\prime }=F(y/x)$ sense.

Apply the integral bounds for a finite harmonic sum to the omitted terms from thirteen through one hundred:Adding the given $H_{12}=86021/27720$ gives $5.15338\dots <H_{100}<5.22348\dots$. Both bounds lie between $5.15$ and $5.25$, so every value they permit rounds to the same two significant figures:The rounding follows from the bounds, without computing all the intervening terms.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

The last two terms in the preceding result equal $v_s$. ThereforeThe left-hand side of the original partial differential equation is consequently $v_{tt}-v_s$, givingIn these variables it has the form of a heat equation, with logarithmic radial coordinate $s$ playing the evolution variable and angular coordinate $t$ playing the spatial variable. If $u$ is single-valued around the whole punctured plane, then $v(s,t+2\pi )=v(s,t)$ as well; that periodicity is a global coordinate condition, not an extra local transformation term.

Differentiate $v_t=y{u}_x-x{u}_y$ with respect to $t$, remembering that its coefficients also depend on $t$. The product rule and multivariable chain rule giveFor a twice continuously differentiable $u$, equality of mixed partial derivatives gives $u_{yx}=u_{xy}$. HenceThe first-order terms come from differentiating $x$ and $y$ themselves; treating $y{\partial }_x-x{\partial }_y$ as an operator with constant coefficients would miss them.

The coordinate partial derivatives are $x_s=-x$, $y_s=-y$, $x_t=y$, and $y_t=-x$. Apply the multivariable chain rule to $v(s,t)=u(x(s,t),y(s,t))$:All partial derivatives of $u$ are evaluated at $(e^{-s}\sin t,e^{-s}\cos t)$. This is a logarithmic polar coordinate change; its Jacobian determinant is $x^2+y^2=e^{-2s}>0$, so it is locally invertible away from the origin. A full angular turn repeats the same point, as usual for polar coordinates.

Rationalize the difference to reveal its small size instead of comparing its two large summands separately:For any fixed real $\alpha$, the right-hand series is a constant multiple of a convergent P-series. The comparison test for series provesThe case $\alpha =0$ is the identically zero series. For nonzero $\alpha$, its summand is asymptotic to ${\alpha }^4/(2n^2)$, explaining the same convergence behavior as the inverse-square series.

For $a_n=n^2{e}^{-\alpha n}$, the ratio test givesThis is less than one exactly when $\alpha >0$. For $\alpha =0$ the terms are $n^2$, and for $\alpha <0$ they grow exponentially; in both cases they fail to tend to zero. Consequently

This is a geometric series with ratio $r=2\alpha /({\alpha }^2+1)$. Since ${\alpha }^2+1>0$, the condition $|r|<1$ is equivalent toIt holds for every real $\alpha$ except $\alpha =\pm 1$. At $\alpha =1$ the terms are all one; at $\alpha =-1$ they alternate between $-1$ and $1$. Neither sequence of terms tends to zero. ThusAll the convergent cases have absolute convergence, including $\alpha =0$, when every term is zero.

For every real $\alpha$, $|\cos (\alpha n)/n^2|\le 1/n^2$. The bounding P-series with exponent two converges. The comparison test for series therefore givesOscillations of cosine are unnecessary for convergence here.