For a continuous random variable, the probability density integrates to the probability of an interval:Individual endpoints have zero probability, so strict or non-strict inequalities give the same answer.

With $\beta =0$, integrating ${\Phi }_x=2\alpha x+{\alpha }^{2}y+\alpha z$ givesThe other two partial derivatives require $K_y=K_z=0$. The origin condition fixes the remaining constant to zero. Hence the scalar potential isIts value at $(1,1,1)$ gives ${\alpha }^2+2\alpha =1$, so

A continuously differentiable conservative vector field has zero curl. HereThis vanishes everywhere if and only if $\boxed{\beta =0}$. Since the field is smooth on the simply connected domain ${\mathbb{R}}^3$, zero curl is also sufficient; the explicit scalar potential in the next part confirms it.

It is helpful first to separate an exact gradient:The fundamental theorem for line integrals makes the gradient contribution $2\alpha +{\alpha }^2$ for either path. On the curved path, $dy=2tdt$, $dz=3t^{2}dt$, $y=t^2$ and $z=t^3$, so the remaining contribution isThus

A line integral uses $\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)$, including the tangent vector. With $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, the three components sum toConsequently

At each regular point, the gradientis normal to the level set and points towards increasing $f$. The arrows in the contour plot above show this direction; their lengths have been normalized to keep the outer arrows legible. Near the origin they point outwards, as expected for a local minimum. Near $(0,1)$, write $\eta =y-1$; the linear approximation is $\nabla f\simeq (3x,-\eta )$. Thus the arrows point away along the horizontal direction and towards the point along the vertical direction. The other two saddle points have the same behaviour after rotations through $2\pi /3$. All four stationary points have zero gradient, so no direction arrow is assigned at the points themselves.

The most useful guide to the contour plot is the common saddle value. Direct polynomial factorization givesThus the level set $f=1/6$ is exactly the three complete linesTheir intersections are the three saddle points, and they enclose the origin in an equilateral triangle. Near the local minimum, the level sets are nearly circles because the leading term is $(x^2+y^2)/2$. For $0<c<1/6$, a closed contour surrounds the origin, while three separate open branches lie outside the triangular barrier. At the saddle level these components meet at the saddle points; above it the central closed contour has opened into the three outer channels.

The identity in polar coordinatesshows the rotation symmetry through $2\pi /3$. It identifies the function as the Hénon-Heiles potential and helps organize all three channels in the original sketch below. The dashed straight lines are the saddle level; labels on the other contours give $f$.

The Hessian matrix isAt the origin it is the identity, so the origin is a strict local minimum. At $(0,1)$ it is $\mathrm{diag}(3,-1)$, so this point is a saddle point. At either remaining stationary point,which again gives a saddle point. All three saddle points have eigenvalues $3$ and $-1$; there are no local maxima.

The gradient vanishes whenIf $x=0$, the second equation gives $y=0$ or $y=1$. If $y=-1/2$, it gives $x^2=3/4$. These exhaust the possibilities. The stationary points and their values are

Let $\mathbf{d}=\mathbf{f}-\mathbf{e}$. A point on the line has position vector $\mathbf{r}=\mathbf{e}+t\mathbf{d}$. By bilinearity and antisymmetry of the cross product,For a point with position vector $\mathbf{c}$, the area of the parallelogram spanned by $\mathbf{c}-\mathbf{e}$ and $\mathbf{d}$ equals base length times perpendicular Euclidean distance. Hence the shortest Euclidean distance to the line isFor $\mathbf{c}=(0,0,1)$, these quantities areThus $\boxed{d=\sqrt{6}/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.

Two directions within the plane areTheir cross product is $(1,1,1)/4$, so a unit vector normal to the plane isIts negative is equally valid. An equation of a plane through three points in vector form isor, parametrically, $\mathbf{r}=\mathbf{e}+s(\mathbf{f}-\mathbf{e})+t(\mathbf{g}-\mathbf{e})$. Since the coordinate sum of $E$ is one, the Cartesian equation of a plane isThe distance from a point to a plane is the absolute normal displacement, giving

The direction must first be represented by a unit vector: $\hat{\mathbf{q}}=(1,1,1)/\sqrt{3}$. The scalar projection of the vector area is thereforeIf a vector projection is intended, multiply this scalar by $\hat{\mathbf{q}}$:

For the outward orientation of a spherical cap, its vector area element isIn the surface integral defining the vector area, the $x$ and $y$ components vanish by integration over a complete azimuth. The remaining component isConsequentlyThis also equals the area of the cap's projection onto the horizontal plane, directed upwards.

Let the semicircle have radius $R$, and write $A=|\mathbf{a}|$, $B=|\mathbf{b}|$. The right triangle has hypotenuse $2R$, so the Pythagorean theorem gives $A^2+B^2=4R^2$. Its area is $AB/2$, whereas half the area of the semicircle is $\pi R^2/4$. Equating these and eliminating $R$ givesEquivalently, the relation between the two norms can be writtenThe two possible ratios are reciprocals, corresponding to interchanging the two sides.

By Thales's theorem, the two sides meeting on the circular arc are orthogonal, because the opposite side is a diameter. Both vectors point towards that meeting point; reversing both does not change their dot product. Consequently

This is a finite geometric series with first term and common ratio both equal to four. Therefore

There are $100$ terms in this arithmetic series, starting at $4$ and ending at $202$. Pairing the first and last terms, the second and penultimate terms, and so on, givesThe summand in the PDF is $2n$, not $2^n$.