Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

Reverse the order in the nonnegative double integral, using $0\le y<\infty$ and $y\le x\le 2y$. For $y>0$, an antiderivative in $x$ is $-(x/y+1/y^2)e^{-xy}$. ThusLet $J$ denote the requested improper integral. Its numerator is $y^2+O(y^4)$ near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integralsIt follows thatThe two terms divided separately by $x^2$ would each have a divergent integral at zero; the cancellation in their difference is essential.

For fixed $x>0$, integrate first over $y$:The integrand is nonnegative, so this order of the double integral is legitimate even before its finiteness is known. Applying the scaled Gaussian integral ${\int }_0^{\infty }{e}^{-a{x}^2}dx=\sqrt{\pi }/(2\sqrt{a})$ yields

The region is an unbounded wedge in the first quadrant, lying between the rays $y=x/2$ and $y=x$. Both boundaries and the origin are included. For integration in the opposite order, the same region is $y\ge 0$ with $y\le x\le 2y$.

This is a Bernoulli differential equation with exponent four. On the nonzero branch selected by $y(0)=1$, put $v=y^{-3}$. Multiplying the original equation by $-3y^{-4}$ givesIts integrating factor is $e^{-x}$, so $(e^{-x}v{)}^{\prime }=-3$ and $e^{-x}v=C-3x$. The initial condition sets $C=1$. ThusThis is the maximal real solution interval containing zero: the solution diverges as $x\to (1/3{)}^{-}$. The real negative branch of the same algebraic formula for $x>1/3$ is disconnected from this initial-value solution. The original equation also admits $y=0$, but it does not satisfy the prescribed initial condition.

For the solution through $x=0$, work on the connected domain $x>-1$. While $y>0$, separation of variables givesThe initial condition gives $C=2$, soThe restriction matters: the unsquared equation requires $1-\frac{1}{2}\ln (1+x)\ge 0$. The solution reaches zero at $x_{\ast }=e^2-1$. Continuing the squared formula beyond that point would give a positive derivative, contradicting the original negative square-root right-hand side.

The nonnegative forward continuation is insteadThe two pieces join with $y^{\prime }=0$, and both satisfy the original equation. Because the right-hand side is nonpositive for $x>-1$, a nonnegative solution cannot leave zero in the forward direction. This is a finite-time extinction phenomenon for a separable differential equation.

The two independently computed answers agree:This verifies the divergence theorem. Its hypotheses hold because the field is smooth and the prism has a closed piecewise smooth boundary with consistently outward normals.

Use the prism bounds read from the PDF diagram and the divergence from part (ii):Set $A=4-2x$ in the inner evaluation. Since $4x-9=-2A-1$, the inner integral is $A^2-(2A+1)A=-A^2-A$. Therefore

Differentiate the corresponding components:

The PDF diagram specifies the prism asIts triangular horizontal cross-section has area four. Calculate the outward flux on each face using its outward normal vector.

On $x=0$, the normal is $-{\mathbf{e}}_x$ and $F_x=0$, so the flux is zero. On $y=0$, the normal is $-{\mathbf{e}}_y$, givingOn the sloping vertical face $y+2x=4$, the outward normal is proportional to $(2,1,0)$. Both $F_x$ and $F_y$ vanish there, while $F_z$ is tangent to the face, so the flux is zero.

On the bottom $z=0$, the normal is $-{\mathbf{e}}_z$ and $F_z=1$, so the flux is minus the base area, namely $-4$. On the top $z=1$, $F_z=0$, so the flux is zero. Thus all five face values and their sum are

The Cartesian curl isEach component measures circulation in the plane perpendicular to its corresponding coordinate axis.

For a Cartesian vector field $\mathbf{F}=(F_x,F_y,F_z)$, its divergence is

Use $u=\cos x$, so $du=-\sin xdx$. Up to an overall minus sign, the differential form becomesSeeking a monomial integrating factor $u^{\alpha }{y}^{\beta }$, equality of mixed derivatives givesThese give $\alpha =1,\beta =0$, so the integrating factor in the original variables isOne may verify exactness directly without relying on invertibility of the cosine substitution:Thus the implicit first integral isThe equivalence with the differential equation holds directly where $\cos x\ne 0$. The zero solution $y=0$ is included at $C=0$. At a zero of $\cos x$, the original graph equation requires $y=0$ for any finite differentiable continuation, so no extra finite graph with nonzero $y$ is created by multiplying by the vanishing factor. If one allows vertical parametrized integral curves of the differential form, $x=\pi /2+n\pi$ is also such a curve and is already contained in the level set $C=0$.

Write the differential form as $Mdx+Ndy$, with the coefficients as printed. A monomial integrating factor can be found from $\mu =x^{\alpha }{y}^{\beta }$. The exactness condition ${\partial }_y(\mu M)={\partial }_x(\mu N)$ givesSolving yields $\alpha =0,\beta =-3$. Thus, on $y\ne 0$,After multiplication, the differential form isHence the implicit solutions areThe branch $\boxed{y=0}$ was excluded by the integrating factor but satisfies the original equation and must be restored. On any graph branch with $y\ne 0$, differentiation of the boxed first integral recovers the original differential equation.

Under independence, there are two disjoint ways to score two goals in total. One game can contain both, with probabilityAlternatively, two different games can contain one goal each, with probabilityAdding givesThis also follows from the Sum of independent Poisson random variables: the season total is Poisson with parameter $10\lambda$, whose probability at two is $e^{-10\lambda }(10\lambda {)}^2/2!$.

Continue with the independent-games assumption from part (i). Exactly one of the ten games must have one goal, while the other nine have zero. The ten choices of the scoring game are disjoint, so

The intended season calculation needs the additional assumption that scores in different games are independent random variables. The printed statement gives their common one-game distribution but does not explicitly state independence; identical marginals alone do not determine season probabilities.

Under independence, each game has probability $1-P(X=0)=1-e^{-\lambda }$ of at least one goal, soWithout independence this product is not determined by the supplied marginal distribution.

The zero-goal term contributes nothing to the expected value. Shift the summation index in the remaining terms:The exponential series justifies the convergent reindexing. Thus the parameter is the mean goals per game.

For a nonnegative integer $K$, add the probabilities of the disjoint events with $0,1,\dots ,K$ goals. The cumulative distribution function is

Normalization of the probability mass function requiresThusThe resulting law is the Poisson distribution with parameter $\lambda$.