Differentiate the ellipsoidal parametrization. Its outward vector surface element isFor $a,b,c>0$, its dot product with $\mathbf{r}$ simplifies to $abc\sin \theta$. Thus the requested direct surface integral isAs a check, the divergence of $\mathbf{r}$ is three, and the ellipsoid has volume $4\pi abc/3$, so the divergence theorem gives the same value.

Take the outward orientation of a surface. The divergence is $\nabla \cdot \mathbf{F}=y^2+z^2+x^2$, so the divergence theorem converts the closed surface flux into an integral over the box:Each term separates; for example the $x^2$ term is $(2a^3/3)(2b)(2c)$. Summing givesHere $a,b,c>0$ are the box half-lengths. Reversing orientation of a surface would negate the result.

Use $u=1/x$, so $dx=-du/u^2$ and $\ln x=-\ln u$. Reversing the integration bounds gives, without evaluating either side,For $J={\int }_0^{\infty }\ln x/(1+x^2)dx$, the same change of variables givesso $\boxed{J=0}$. These are legitimate transformations of improper integrals: $|\ln x|$ is integrable near zero, and $(\ln x)/x^2$ is integrable at infinity. In particular $J$ is absolutely convergent, so no cancellation of divergent endpoints is being assumed.

For $f=x\ln x$, the derivatives are $f^{\prime }=1+\ln x$ and $f^{\prime \prime }=1/x>0$. Its unique stationary point is the global minimum $(e^{-1},-e^{-1})$. For $g=\ln x/x$, $g^{\prime }=(1-\ln x)/x^2$ changes from positive to negative at $x=e$, giving the global maximum $(e,e^{-1})$. Their endpoint limits areAn intersection requires $(x^2-1)\ln x=0$, and on $x>0$ both factors vanish only at $x=1$. There $f=g=0$ and $f^{\prime }=g^{\prime }=1$, proving a common tangent line $y=x-1$. In fact,because the two numerator factors always have the same sign. Thus the curves touch at $(1,0)$ without exchanging their order.

The fundamental theorem of calculus and part (a) giveThe difference is constant in $\theta$. Since $I(0,k)=E(0,k)=0$, the constant is zero, provingThis is simply the chain rule integrated, with $E$ itself as the antiderivative of the square-root factor.

Work first with $|k|<1$, so the real integrands and their parameter derivatives are smooth on every finite amplitude interval. For the elliptic integral of the second kind,On the other hand, subtracting the integrands of the second-kind and first-kind elliptic integrals givesConsequentlyThis is the modulus derivative of the incomplete elliptic integral of the second kind. At $k=0$ its right-hand side has a removable limit zero, agreeing with $E_k(\theta ,0)=0$. Both $E$ and $F$ here use amplitude $\theta$ and modulus $k$; differentiating with respect to the alternative parameter $k^2$ would give a different prefactor.

The fundamental theorem of calculus differentiates the upper endpoint, while differentiation under the integral sign differentiates the parameter:These formulas hold, for example, when $g$ and its displayed parameter partial derivative are continuous on a neighborhood of the integration interval and the parameter value.

If $Av=\lambda v$, then $A^{2}v={\lambda }^{2}v$. Alternatively, direct matrix multiplication givesThus the normalized eigenvectors $v_{+},v_{-}$ from part (b) have respective eigenvaluesIf $\sin 2\theta =0$, then $B=I$, and the whole two-dimensional space is its eigenspace. This degeneracy includes cases where the two eigenvalues of $A$ were distinct but opposite.

Adding and subtracting the two coordinates shows that $A(1,1{)}^T=(c+s)(1,1{)}^T$ and $A(1,-1{)}^T=(c-s)(1,-1{)}^T$. The normalized eigenvectors can therefore be chosen asTheir inner product is $(1-1)/2=0$, and their squared norms are one, proving orthonormality. If $s=0$, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

Put $c=\cos \theta$, $s=\sin \theta$. The characteristic polynomial is $(c-\lambda {)}^2-s^2$, givingWhen $s=0$, these eigenvalues coincide and $A=cI=\pm I$.

Use the normal vector $\mathbf{N}=(-8,-3,-9)$ found from the edge vectors, rather than the normal vector to the origin-based linear span in part (b). The required plane satisfies $\mathbf{N}\cdot (\mathbf{r}-\mathbf{a})=0$. Since $\mathbf{N}\cdot \mathbf{a}=5$,After dividing by $|\mathbf{N}|=\sqrt{154}$, its equation has a unit normal. Its perpendicular distance from the origin is thus $\boxed{5/\sqrt{154}}$.

The two edge vectors based at $A$ are $\mathbf{b}-\mathbf{a}=(-3,-1,3)$ and $\mathbf{c}-\mathbf{a}=(-3,2,2)$. Their cross product is $(-8,-3,-9)$, with squared Euclidean norm $64+9+81=154$. The triangle occupies half the corresponding parallelogram, so

A normal vector to the plane spanned by $\mathbf{a},\mathbf{b}$ is their cross product:Dividing by this Euclidean norm gives the two possible choices of unit normal,

Subtract the two position vectors to obtain the direction $\mathbf{b}-\mathbf{a}=(-3,-1,3)$. The affine line in a vector space through the points is thereforeThe values $t=0,1$ recover the two specified points.