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The gradient vanishes when $3x^2-2=3y^2-2=0$. Put $s=\sqrt{2/3}$. The Hessian matrix is $\mathrm{diag}(6x,6y)$, so the second-derivative test gives:

The positive-definite and negative definite cases give small closed level curves around the minimum and maximum. At each saddle, the zero contour crosses itself. In fact,so the entire zero level curve is the union of the line $y=-x$ and the ellipse $x^2-xy+y^2=2$. This factorization fixes the topology of a rough contour sketch. Neither local extremum is global, since the cubic is unbounded above and below.

Construct a global vector-field potential rather than relying only on vanishing curl:Thus $\mathbf{F}$ is a conservative vector field. The endpoints of the given parametric curve are $\mathbf{r}(0)=(3,0,0)$ and $\mathbf{r}(\pi /2)=(0,2,1)$. The fundamental theorem for line integrals givesThe intervening turns of the curve have no effect because this line integral is path independent.

The first-order multivariable Taylor expansion isTaking $(x_0,y_0)=(2,0)$ gives the linear approximationTo assess its significant figures, the quadratic Taylor polynomial supplies the correction $h^2+4hk+2k^2=-0.0001$ for $h=0.03$, $k=-0.01$. Direct evaluation gives $2.03^2{e}^{-0.01}=4.0798963599\dots$. Thus the estimate is correct to four significant figures: both round to $4.080$, whereas to five significant figures the exact value is $4.0799$ and the approximation is $4.0800$. The unexpectedly small quadratic correction results partly from cancellation of its mixed and unmixed terms.

The divergence theorem includes every part of the closed boundary, including the two vertical cut faces. Its volume integrand simplifies because the terms linear in $x$ cancel:In cylindrical coordinates, the remaining integration region is $0\le r\le 1$, $0\le z\le 1-r^2$, $\pi /2\le \theta \le 2\pi$. With the Jacobian determinant $r$, the outward flux isThe angular range is three quarters of a full revolution. The volume calculation already accounts for the base and cut faces, so no additional surface terms should be appended.

Use implicit differentiation while holding $x$ fixed. Writing $y_z=(\partial y/\partial z{)}_x$, the constraint givesThe chain rule applied to $v=x{y}^3$ now yieldsThis local formula requires $2xy+z^2\ne 0$, as in the implicit function theorem; that denominator equals $3$ at the specified point.

The arc length of this parametric curve is the integral of its speed. Differentiation gives $x^{\prime }(t)=at\cos t$ and $y^{\prime }(t)=at\sin t$, soFor a positive scale parameter this is $a/2$. The absolute value is necessary when $a$ is allowed to be any real constant; $a=0$ gives a degenerate curve of zero length. The nondegenerate curve is an involute of a circle.