An integrating factor isMultiplying the linear ordinary differential equation by it givesThe right-hand side integrates to $(a^2+x^2{)}^{3/2}/3$, soThis holds on every interval where the coefficient is defined. If $a=0$, intervals cannot cross $x=0$; equivalently the family is $y=x^2/3+C/x$ on either half-line, with an independently chosen constant.

Cubing and adding the parametrized coordinates yieldsTherefore the implicit algebraic curve isFor $a\ne 0$, every nonorigin real point on this curve has $x\ne 0$ and can be recovered using $t=y/x$; the origin is already included. The equation is invariant under exchanging $x$ and $y$, so its reflection line is $\boxed{y=x}$. In the parametrization that reflection corresponds to replacing $t$ by $1/t$. If $a=0$, the implicit cubic degenerates to the line $x+y=0$, whereas the given parametrization only describes the origin, so this degenerate case must be kept separate.

Take $a>0$ for the sketch; changing its sign rotates the picture by $\pi$. The parametric curve passes through $(0,0)$ at $t=0$ and through $(3a/2,3a/2)$ at $t=1$. As $t\to \pm \infty$ it returns towards the origin, since $x\sim 3a/t^2$ and $y\sim 3a/t$. The positive-$t$ branch makes a loop in the first quadrant, leaving the origin tangent to the positive $x$ axis and returning tangent to the positive $y$ axis.

For a more precise sketch, differentiation givesThe loop has its largest $x$ at $t=2^{-1/3}$ and largest $y$ at $t=2^{1/3}$. Their coordinates are $a(2^{2/3},2^{1/3})$ and its reflection in $y=x$.

The parameter value $t=-1$ is excluded. Cancelling the common factor in the sum givesThus the asymptote is $\boxed{x+y+a=0}$. As $t\to -1^{-}$, $x\to +\infty$ and $y\to -\infty$; as $t\to -1^{+}$, $x\to -\infty$ and $y\to +\infty$. The branch $t<-1$ approaches the origin from the fourth quadrant as $t\to -\infty$, while $-1<t<0$ connects the other infinite branch to the origin through the second quadrant.

This is the folium of Descartes. If $a=0$, the parametrization degenerates to the origin for every defined parameter value and there is no nontrivial loop.

The upward orientation of ${\mathbf{X}}_r\times {\mathbf{X}}_{\theta }$ corresponds to traversing the boundary from the apex to the starting rim, around the rim in increasing azimuth, and back to the apex. This is the indicated $ABCA$ orientation and is outward from the solid. Integrate that vector area element directly:The radial integrals are $2aR/3$ in the horizontal components and $b^2/2$ in the vertical component. Hence the vector area isThe absolute value is needed only because the original statement permits negative $b$. Reversing the boundary orientation reverses the entire vector. For a full rotation the horizontal components cancel, leaving $\pi b^2\mathbf{k}$, the signed projected area of the base.

For $a/|b|\ll 1$, apply the binomial series to the result in (c):ThereforeFor $\alpha =2\pi$, the leading term is $\pi b^2$, the area of the base disk. A very shallow cap is nearly planar; its first area excess is $\pi a^2$.

Use the embedded surface parametrizationIts tangent vectors have cross productwhose magnitude is $r\sqrt{1+4a^2{r}^2/b^4}$. Thus the surface area isWith $u=1+4a^2{r}^2/b^4$, the radial integral becomes $b^4(8a^2{)}^{-1}\int u^{1/2}du$. Evaluating its endpoints givesThis is the scalar-area part of the paraboloid sector surface measures.

The volume element in cylindrical coordinates is $rdzdrd\theta$. Integrating from the base to the curved surface givesHenceFor a complete rotation this is half the volume of the enclosing cylinder.

Put $R=|b|$ so that the radial range respects the stated allowance $b\ne 0$. With the azimuth measured from the starting meridian, the cylindrical coordinates of the curved surface satisfyAt $r=0$ all azimuths describe the same apex. The surface is a sector of an elliptic paraboloid.

The opposite face lies in the plane spanned by $\mathbf{a},\mathbf{b}$ through the origin. Directly,The distance from a point to a plane is the absolute normal component, soHere $\mathbf{c}=(\mathbf{a}\times \mathbf{b})/2$, so the perpendicular foot is the origin, a vertex of the opposite triangular face. Thus the distance to the actual face equals the distance to its containing plane.

Set $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. Since these vectors form a basis, write $\mathbf{r}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. The reciprocal basis givesA point is in the closed tetrahedron precisely when it is a convex combination of the four vertices, with weights $1-\alpha -\beta -\gamma ,\alpha ,\beta ,\gamma$. Thus the necessary and sufficient conditions areThese four half-spaces are the four inward sides of its faces. This is tetrahedron membership via scalar triple products; making all inequalities strict describes the interior.

Take two edge vectors in the face, $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product is the face normal vectorThe face plane is $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Every term in $\mathbf{a}\cdot \mathbf{n}$ except $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ vanishes, givingThe scalar triple product is nonzero, so this normal cannot vanish.

The gradient vanishes at the origin. Its Hessian matrix there iswith eigenvalues $1$ and $-1$. The quadratic term is therefore indefinite, so the origin is a saddle point of a scalar function. Equivalently, the leading term $xy$ is positive along $y=x$ and negative along $y=-x$.