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The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.

Because the equation depends on the ratio $y/x$, set $u=y/x$, so $y^{\prime }=u+x{u}^{\prime }$. Dividing by $x^2$ and substituting givesThis reduces the homogeneous first-order differential equation to a separable differential equation. For $u\ne 1$,At $x=e$, the condition $y=0$ gives $u=0$, so $1=1+C$ and $C=0$. HenceThe interval displayed is the maximal solution interval containing $x=e$: there is a pole at $x=1$. Separating variables omitted the solution $u=1$, or $y=x$, but that solution does not satisfy the given condition.

Apply the same dimensionless transformation independently to the two cups, giving $X,Y\in [-\pi /2,\pi /2]$. The total-volume condition becomes $X+Y\ge \pi /2$. Since neither variable can exceed $\pi /2$, this region has $0\le X\le \pi /2$ and $\pi /2-X\le Y\le \pi /2$, the triangle shown above.

By independence of random variables, the joint probability density is $\frac{1}{4}\cos x\cos y$. The desired double integral isThusBoundary equality has probability zero for these continuous random variables.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

The upward unit vector extracts the vertical curl component:It is constant over the unit disc, soThis agrees with part (d) by Stokes theorem: the boundary circle has counterclockwise orientation when viewed from the upward normal, including for the parametrizations $2\pi t^n$.

The original PDF has angle $\theta (t)=2\pi t^n$, with $n$ an exponent on $t$. It is not $2\pi nt$: each permitted value of $n$ changes the speed but traverses the same circle once counterclockwise. Along the path, $r^2=2$ andThe radial contribution to the line integral vanishes, since $(\cos \theta ,\sin \theta ,1)\cdot (-\sin \theta ,\cos \theta ,0)=0$. The rotational contribution givesThereforeThe answer is independent of the monotone parametrization, as an oriented line integral should be.