past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperia_3_2021.pdf

= 1D
{parent=Paper 3}
{scope}
{title2=Groups}

= i
{parent=1d}
{scope}

= Solution
{parent=i}

Suppose $G/Z(G)$ is generated by $gZ(G)$. Every element of $G$ can then be written $g^az$ with $z\in Z(G)$. Hence
$$
(g^az_1)(g^bz_2)=g^{a+b}z_1z_2
=(g^bz_2)(g^az_1).
$$
Thus the <cyclic quotient by the center> condition forces $G$ to be abelian.

If $|H/Z(H)|=7$, the quotient has prime order and is therefore cyclic. The result just proved would make $H$ abelian, so $Z(H)=H$ and the quotient would have order one, a contradiction. Therefore
$$
\boxed{\text{no such group }H\text{ exists}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=1d}
{scope}

= Solution
{parent=ii}

Such a group does exist. Take $H=S_3$. Its center is trivial, so
$$
H/Z(H)\cong S_3
$$
has order six. Therefore
$$
\boxed{H=S_3\text{ is an example}.}
$$

Solved by gpt-5.6-sol high.

= 2D
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=2D}

Elements $g,h\in G$ are conjugate when
$$
h=xgx^{-1}
$$
for some $x\in G$. Then
$$
h^n=xg^nx^{-1}
$$
for every integer $n$, so $h^n=e$ exactly when $g^n=e$. They therefore have the same <order of a group element>[order].

The Möbius group consists of the <Möbius transformation>[Möbius transformations]
$$
z\longmapsto\frac{az+b}{cz+d},
\qquad ad-bc\ne0,
$$
of the Riemann sphere, under composition. If $h=xgx^{-1}$, then
$$
h(x(p))=x(g(p)).
$$
Thus $p$ is fixed by $g$ exactly when $x(p)$ is fixed by $h$. Conjugation gives a bijection between their fixed-point sets, so conjugate elements have the same number of fixed points.

A nonidentity Möbius transformation has at most two fixed points because its fixed-point equation is quadratic on the Riemann sphere. If it has only one repeated fixed point, conjugate that point to infinity; the transformation becomes a nontrivial translation $z\mapsto z+c$, which has infinite order. Consequently every nontrivial finite-order element has two distinct fixed points:
$$
\boxed{|\operatorname{Fix}(g)|=2}.
$$
This is <fixed points of a finite-order Möbius transformation>.

Solved by gpt-5.6-sol high.

= 3B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=3b}
{scope}

= Solution
{parent=a}

In <Einstein notation>,
$$
\begin{aligned}
[\nabla\times(\psi A)]_i
&=\epsilon_{ijk}\partial_j(\psi A_k)\\
&=\psi\epsilon_{ijk}\partial_jA_k
+\epsilon_{ijk}(\partial_j\psi)A_k.
\end{aligned}
$$
This is
$$
\boxed{\nabla\times(\psi A)
=\psi\nabla\times A+\nabla\psi\times A}.
$$

Similarly,
$$
\begin{aligned}
\nabla\cdot(A\times B)
&=\partial_i(\epsilon_{ijk}A_jB_k)\\
&=\epsilon_{ijk}(\partial_iA_j)B_k
+\epsilon_{ijk}A_j(\partial_iB_k)\\
&=B\cdot(\nabla\times A)-A\cdot(\nabla\times B),
\end{aligned}
$$
after cyclically relabelling the first term and swapping two indices in the second. Hence
$$
\boxed{\nabla\cdot(A\times B)
=B\cdot\nabla\times A-A\cdot\nabla\times B}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=3b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The data are rotationally symmetric, so seek a radial solution. The <radial Laplacian> in two dimensions gives
$$
\frac1r(ru')'=16r^2.
$$
Twice integrating,
$$
u(r)=r^4+C\log r+D.
$$
Regularity at the center forces $C=0$, and $u(1)=1$ gives $D=0$. Thus
$$
\boxed{u(r)=r^4}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

On the annulus the logarithmic term is allowed, so
$$
u(r)=r^4+C\log r+D.
$$
The condition $u(1)=1$ gives $D=0$, while $u(2)=1$ gives
$$
16+C\log2=1.
$$
Therefore
$$
\boxed{
u(r)=r^4-\frac{15\log r}{\log2}}.
$$

Solved by gpt-5.6-sol high.

= 4B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=4b}
{scope}

= Solution
{parent=a}

A second-rank tensor is antisymmetric when
$$
T_{ij}=-T_{ji}.
$$
Under an orthogonal Cartesian coordinate change with matrix $R$,
$$
T'_{ij}=R_{ik}R_{j\ell}T_{k\ell}.
$$
Therefore
$$
\begin{aligned}
T'_{ji}
&=R_{jk}R_{i\ell}T_{k\ell}\\
&=-R_{jk}R_{i\ell}T_{\ell k}\\
&=-T'_{ij},
\end{aligned}
$$
after relabelling the dummy indices. Hence the <antisymmetric second-rank tensor> property is independent of Cartesian coordinates.

Solved by gpt-5.6-sol high.

= b
{parent=4b}
{scope}

= Solution
{parent=b}

For $F=(y,z,x)$,
$$
T_{ij}=\frac{\partial F_i}{\partial x_j}
=
\begin{pmatrix}
0&1&0\\
0&0&1\\
1&0&0
\end{pmatrix}.
$$
Its antisymmetric part is
$$
\boxed{
\frac12(T-T^T)=
\begin{pmatrix}
0&1/2&-1/2\\
-1/2&0&1/2\\
1/2&-1/2&0
\end{pmatrix}}.
$$
Also,
$$
B=\nabla\times F=(-1,-1,-1).
$$
For example, $-(1/2)\epsilon_{12k}B_k=1/2$, and the same componentwise calculation gives
$$
\boxed{
\frac12(T_{ij}-T_{ji})
=-\frac12\epsilon_{ijk}B_k}.
$$

Solved by gpt-5.6-sol high.

= 5D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=5d}
{scope}

= Solution
{parent=a}

The order of an element $x$ is the least positive integer $d$ such that $x^d=e$, and the order of a finite group is its number of elements.

<Lagrange theorem> states that if $H\leq G$ and $G$ is finite, then
$$
|H|\mid |G|,
\qquad
|G|=[G:H]|H|.
$$
Indeed, the left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection from $H$ to each coset, so every coset has $|H|$ elements. Summing over the $[G:H]$ cosets proves the formula.

The cyclic subgroup $\langle x\rangle$ has exactly $d$ elements when $x$ has order $d$. Applying Lagrange to $\langle x\rangle\leq G$ gives
$$
\boxed{\operatorname{ord}(x)\mid |G|}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=5d}
{scope}

= Solution
{parent=b}

No. The alternating group $A_4$ has order twelve, and six is a proper divisor of twelve, but $A_4$ has no element of order six. Its nonidentity elements are eight 3-cycles of order three and three double transpositions of order two.

Solved by gpt-5.6-sol high.

= c
{parent=5d}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Let $a$ and $b$ generate $C_m$ and $C_n$. The order of $(a,b)$ in the <direct product of groups> is
$$
\operatorname{lcm}(m,n).
$$
When $\gcd(m,n)=1$, this is $mn=|C_m\times C_n|$. Hence $(a,b)$ generates the whole product and
$$
\boxed{C_m\times C_n\cong C_{mn}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Every element satisfies $x^2=e$, including the identity. For any $x,y\in G$,
$$
e=(xy)^2=xyxy.
$$
Since $x^{-1}=x$ and $y^{-1}=y$, this implies $xy=yx$, so $G$ is abelian. It is therefore a vector space over the two-element field, with group operation as vector addition and scalar multiplication by zero or one. A basis of this finite vector space gives
$$
\boxed{G\cong C_2^r
=\underbrace{C_2\times\cdots\times C_2}_{r\text{ factors}}}
$$
for some $r\geq0$.

Solved by gpt-5.6-sol high.

= d
{parent=5d}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

Write $d=qr+s$ with $0\leq s<r$. Since $x^d=e\in H$ and $x^r\in H$,
$$
x^s=x^d(x^r)^{-q}\in H.
$$
The minimality of the positive integer $r$ forces $s=0$. Therefore
$$
\boxed{r\mid d}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

Among the $n+1$ right cosets
$$
H,Hx,Hx^2,\ldots,Hx^n
$$
two must coincide because $[G:H]=n$. If $Hx^i=Hx^j$ with $0\leq i<j\leq n$, right multiplication by $x^{-i}$ gives $H=Hx^{j-i}$, so
$$
x^{j-i}\in H,
\qquad
0<j-i\leq n.
$$

The least such exponent need not divide $n$. For example, take $G=S_3$, $H=\langle(12)\rangle$, and $x=(13)$. Then $[G:H]=3$, while $x\notin H$ and $x^2=e\in H$, so the least exponent is two, which does not divide three.

Solved by gpt-5.6-sol high.

= 6D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=6d}
{scope}

= Solution
{parent=a}

For $x\in X$,
$$
\operatorname{Orb}(x)=\{gx:g\in G\},
\qquad
\operatorname{Stab}(x)=\{g\in G:gx=x\}.
$$
The stabilizer contains the identity; if $g,h$ fix $x$, then
$$
(gh^{-1})x=g(h^{-1}x)=gx=x,
$$
so it is a subgroup.

The <orbit-stabilizer theorem> states
$$
\boxed{|G|=|\operatorname{Orb}(x)|\,|\operatorname{Stab}(x)|}.
$$
Define
$$
G/\operatorname{Stab}(x)\longrightarrow\operatorname{Orb}(x),
\qquad
g\operatorname{Stab}(x)\longmapsto gx.
$$
The map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

Solved by gpt-5.6-sol high.

= b
{parent=6d}
{scope}

= Solution
{parent=b}

The orbit of $(1,\ldots,k)$ consists precisely of the ordered $k$-tuples of distinct elements of $\{1,\ldots,n\}$:
$$
\operatorname{Orb}(x)
=\{(x_1,\ldots,x_k):x_i\ne x_j\text{ for }i\ne j\}.
$$
It has size
$$
n(n-1)\cdots(n-k+1)=\frac{n!}{(n-k)!}.
$$
The stabilizer consists of permutations fixing $1,\ldots,k$ pointwise, while freely permuting the remaining letters:
$$
\operatorname{Stab}(x)\cong S_{n-k},
\qquad
|\operatorname{Stab}(x)|=(n-k)!.
$$
Thus
$$
|\operatorname{Orb}(x)|\,|\operatorname{Stab}(x)|
=\frac{n!}{(n-k)!}(n-k)!=n!=|S_n|,
$$
verifying orbit-stabilizer.

Solved by gpt-5.6-sol high.

= c
{parent=6d}
{scope}

= Solution
{parent=c}

Choose $h\in H\setminus\operatorname{Stab}(x)$ and put $y=hx\ne x$. For any $z\ne x$, double transitivity supplies an element $s\in G$ taking the ordered pair $(x,y)$ to $(x,z)$. Since $sx=x$, one has
$$
s\in\operatorname{Stab}(x)\subseteq H.
$$
Then $sh\in H$ sends $x$ to $z$. Together with the identity for $z=x$, this proves that $H$ acts transitively on $X$.

Now let $g\in G$. Transitivity of $H$ gives $h_0\in H$ with $h_0x=gx$. Hence
$$
h_0^{-1}g\in\operatorname{Stab}(x)\subseteq H,
$$
so $g\in H$. Therefore
$$
\boxed{H=G}.
$$
This is the point-stabilizer maximality property of a <doubly transitive group action>.

Solved by gpt-5.6-sol high.

= 7D
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=7D}

Let $G$ act on its underlying set by left multiplication. The resulting homomorphism
$$
\lambda:G\longrightarrow S_G\cong S_n,
\qquad
\lambda_g(x)=gx,
$$
is injective because $\lambda_g(e)=g$. This is <Cayley theorem>. If $g\ne e$, then $gx=x$ would imply $g=e$ after right cancellation, so every nonidentity permutation in the image is fixed-point free.

If $|G|$ is even, pair every element with its inverse. Elements not equal to their inverses occur in pairs. Since the identity is self-inverse and the group has even size, there must be another self-inverse element $g\ne e$. It has order two.

A permutation is odd when its <sign homomorphism>[sign] is $-1$. If a subgroup $H\leq S_m$ contains an odd element, the restriction
$$
\operatorname{sgn}:H\to\{1,-1\}
$$
is surjective. Its kernel is the set of even elements and has index two. Each coset has the same size, so precisely half of $H$ is odd.

Now let $n=4k+2$. An element of order two in the fixed-point-free regular representation is a product of
$$
\frac n2=2k+1
$$
disjoint transpositions, so it is odd. The image $H\cong G$ therefore has an index-two normal subgroup of even permutations. Its order is $2k+1>1$, so it is nontrivial and proper. Hence
$$
\boxed{G\text{ is not simple}}.
$$

Nonabelian simple groups of even order do exist; the smallest example is the alternating group $A_5$, of order sixty.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=8d}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Identify $A$ with $\{(a,1):a\in A\}$. Componentwise use of the stated multiplication shows that this set contains the identity and is closed under products and inverses, so it is a subgroup. It has the two cosets
$$
A,\qquad A(e,-1),
$$
and therefore has index two. Every index-two subgroup is normal, so
$$
\boxed{A\triangleleft D(A)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For $\epsilon=-1$,
$$
(a,-1)^2=(aa^{-1},1)=(e,1),
$$
so every element outside $A$ has order two.

The element $(e,-1)$ commutes with $(a,1)$ exactly when
$$
(a,-1)=(a^{-1},-1),
$$
or $a^2=e$. Since $A$ itself is abelian, $D(A)$ is abelian exactly when every $a\in A$ satisfies $a^2=e$.

Suppose $D(A)$ is nonabelian. An element $(a,1)$ is central exactly when $a=a^{-1}$, equivalently $a^2=e$. No element $(a,-1)$ is central, because commuting with every $(b,1)$ would require $b^2=e$ for every $b$, which would make the whole group abelian. Hence
$$
\boxed{
Z(D(A))=\{(a,1):a^2=e\}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=8d}
{scope}

= Solution
{parent=b}

If $Q\in O(2)$, then $Q^TQ=I$, so
$$
(\det Q)^2=1
$$
and $\det Q=\pm1$. If $\det Q=1$, its first column is $(\cos\theta,\sin\theta)^T$ for some $\theta$, and orthonormality plus positive orientation force
$$
Q=
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta
\end{pmatrix}.
$$
Thus every element of $SO(2)$ is a rotation.

Let $J=\operatorname{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and
$$
\boxed{O(2)=SO(2)\sqcup SO(2)J}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=8d}
{scope}

= Solution
{parent=c}

The standard isomorphisms are
$$
SO(2)\longrightarrow B,
\qquad
\begin{pmatrix}\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\end{pmatrix}
\longmapsto e^{i\theta},
$$
and
$$
\mathbb R/\mathbb Z\longrightarrow B,
\qquad
t+\mathbb Z\longmapsto e^{2\pi it}.
$$

For $z\in B$, let $R_z$ be the corresponding rotation. The identity
$$
JR_zJ=R_{z^{-1}}
$$
shows that
$$
(z,\epsilon)\longmapsto
\begin{cases}
R_z,&\epsilon=1,\\
R_zJ,&\epsilon=-1
\end{cases}
$$
respects the multiplication of the <generalized dihedral group>. The decomposition in part (b) makes it bijective. Hence
$$
\boxed{O(2)\cong D(B)}.
$$

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=9b}
{scope}

= Solution
{parent=a}

For a regular parametrized curve $r(t)$, the <Frenet frame> is
$$
\boxed{
T=\frac{r'}{|r'|},\qquad
B=\frac{r'\times r''}{|r'\times r''|},\qquad
N=B\times T}.
$$
Equivalently, wherever $dT/dt\ne0$, one may write $N=(dT/dt)/|dT/dt|$ and $B=T\times N$.

Solved by gpt-5.6-sol high.

= b
{parent=9b}
{scope}

= Solution
{parent=b}

Differentiation gives
$$
r'(t)=3\sin t\cos t
\left(-2\cos t,\sin t,\sqrt3\sin t\right),
$$
and $|r'|=6|\sin t\cos t|$. Thus
$$
T=\frac{\sigma}{2}
\left(-2\cos t,\sin t,\sqrt3\sin t\right),
\qquad
\sigma=\operatorname{sgn}(\sin t\cos t).
$$
The plus sign applies on $(0,\pi/2)\cup(\pi,3\pi/2)$ and the minus sign on $(\pi/2,\pi)\cup(3\pi/2,2\pi)$.

On each open quadrant,
$$
\boxed{
N=\sigma
\left(\sin t,\frac12\cos t,\frac{\sqrt3}{2}\cos t\right)},
\qquad
\boxed{
B=\left(0,\frac{\sqrt3}{2},-\frac12\right)}.
$$
The constant binormal proves that the curve is planar; indeed $z=\sqrt3y$.

Solved by gpt-5.6-sol high.

= c
{parent=9b}
{scope}

= Solution
{parent=c}

For $n=(a,b,c)$ and $r=(x,y,z)$, the integrand is $\frac12(n\times r)\cdot dr$. Since
$$
\nabla\times\left[\frac12(n\times r)\right]=n,
$$
the <Stokes theorem> gives the signed planar area, and its absolute value gives the ordinary area.

The $xy$ projection of the given curve is the astroid $x=2\cos^3t$, $y=\sin^3t$, whose area is
$$
\frac12\oint(x\,dy-y\,dx)=\frac{3\pi}{4}.
$$
The plane $z=\sqrt3y$ stretches area from its $xy$ projection by $\sqrt{1+3}=2$. Hence
$$
\boxed{\operatorname{Area}=\frac{3\pi}{2}}.
$$

Solved by gpt-5.6-sol high.

= 10B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=10b}
{scope}

= Solution
{parent=a}

Let
$$
I=\int_0^\infty e^{-ax^2}\,dx.
$$
Then
$$
I^2=\int_0^\infty\int_0^\infty
e^{-a(x^2+y^2)}\,dx\,dy.
$$
Using <polar coordinates> on the first quadrant,
$$
I^2
=\int_0^{\pi/2}\int_0^\infty e^{-ar^2}r\,dr\,d\theta
=\frac{\pi}{4a}.
$$
Since $I>0$, the <Gaussian integral> is
$$
\boxed{
\int_0^\infty e^{-ax^2}\,dx
=\sqrt{\frac{\pi}{4a}}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=10b}
{scope}

= Solution
{parent=b}

For fixed $x>0$, $x\ne1$,
$$
\int_0^1x^y\,dy=\frac{x-1}{\log x},
$$
with limiting value one at $x=1$. Integrating over $0<x<1$ in the opposite order gives
$$
\int_0^1\int_0^1x^y\,dy\,dx
=\int_0^1\frac{dy}{y+1}
=\log2.
$$
On the other hand, after setting $x=e^{-u}$,
$$
\int_0^1\frac{x-1}{\log x}\,dx
=\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du.
$$
Consequently
$$
\boxed{
\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du=\log2}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=10b}
{scope}

= Solution
{parent=c}

In <polar coordinates>, the circle is $r=2$, the line $y=\sqrt3x$ is $\theta=\pi/3$, and $y=1$ is $r\sin\theta=1$. The inequalities select
$$
\boxed{
\frac\pi6<\theta<\frac\pi3,\qquad
\csc\theta<r<2}.
$$
The surface height is
$$
z=\frac{xy}{x^2+y^2}
=\sin\theta\cos\theta.
$$
Thus the required volume is
$$
\begin{aligned}
V
&=\int_{\pi/6}^{\pi/3}
\int_{\csc\theta}^{2}
\sin\theta\cos\theta\,r\,dr\,d\theta\\
&=\int_{\pi/6}^{\pi/3}
\left(2\sin\theta\cos\theta-\frac12\cot\theta\right)d\theta\\
&=\boxed{\frac12-\frac14\log3}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= 11B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=11b}
{scope}

= Solution
{parent=a}

Set
$$
x=au,\qquad y=bv,\qquad z=cw.
$$
The ellipsoid becomes the unit ball and the <Jacobian determinant> is $abc$. Therefore
$$
\boxed{
\operatorname{Vol}
=abc\operatorname{Vol}(B_1)
=\frac{4\pi abc}{3}},
$$
assuming $a,b,c>0$.

Solved by gpt-5.6-sol high.

= b
{parent=11b}
{scope}

= Solution
{parent=b}

Fix the first index $i$ and form the vector field with components $F_j=T_{ij}$. The <divergence theorem> gives
$$
\int_SF_jn_j\,dS
=\int_V\partial_jF_j\,dV.
$$
Substitution yields the <tensor divergence theorem>
$$
\boxed{
\int_ST_{ij}n_j\,dS
=\int_V\frac{\partial T_{ij}}{\partial x_j}\,dV}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=11b}
{scope}

= Solution
{parent=c}

For $u=(-y,x,\alpha)$, one has $\nabla\cdot u=0$, and therefore
$$
a_i=\partial_j(u_iu_j)
=(u\cdot\nabla)u_i
=(-x,-y,0)_i.
$$

Close the upper half-ellipsoid with its unit-disk base $D$ in the plane $z=0$. By symmetry,
$$
\int_V(-x,-y,0)\,dV=0.
$$
On the base, the outward normal is $n=(0,0,-1)$, so
$$
u_iu_jn_j=-\alpha u_i
=(\alpha y,-\alpha x,-\alpha^2)_i.
$$
The first two components integrate to zero over the disk, while the third integrates to $-\pi\alpha^2$. The closed-surface tensor identity therefore gives
$$
\int_Su_iu_jn_j\,dS
=(0,0,\pi\alpha^2)_i.
$$
Multiplying by $\beta^2$,
$$
\boxed{F=(0,0,\pi\alpha^2\beta^2)}.
$$

Solved by gpt-5.6-sol high.

= 12B
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=12b}
{scope}

= Solution
{parent=a}

The vector identity from Question 3 gives
$$
\begin{aligned}
\nabla\cdot(E\times B)
&=B\cdot(\nabla\times E)-E\cdot(\nabla\times B)\\
&=-B\cdot\frac{\partial B}{\partial t}
-E\cdot J-E\cdot\frac{\partial E}{\partial t}\\
&=-E\cdot J
-\frac{\partial}{\partial t}
\frac{|E|^2+|B|^2}{2}.
\end{aligned}
$$
Integrating over $V$ and applying the <divergence theorem> proves the <Poynting theorem>
$$
\boxed{
\int_SP\cdot dS
=-\int_VE\cdot J\,dV
-\frac{\partial}{\partial t}
\int_V\frac{|E|^2+|B|^2}{2}\,dV}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=12b}
{scope}

= Solution
{parent=b}

Both divergences vanish, so $\rho=0$. Faraday's and Ampere's curl equations respectively require
$$
kE_0=\omega B_0,
\qquad
kB_0=\omega E_0.
$$
For positive constants these imply
$$
\boxed{\omega=k,\qquad E_0=B_0,\qquad\rho=0}.
$$

The <Poynting vector> is
$$
P=E_0^2\widehat x\cos^2(kx-\omega t).
$$
Only the two $x$-faces of the box contribute to its outward flux. With $a=\pi/(2k)$,
$$
\begin{aligned}
\int_SP\cdot dS
&=L^2E_0^2[
\cos^2(ka-\omega t)-\cos^2(\omega t)]\\
&=-L^2E_0^2\cos(2\omega t).
\end{aligned}
$$
The electromagnetic energy inside the box is
$$
U(t)=L^2E_0^2\int_0^a\cos^2(kx-\omega t)\,dx,
$$
and direct differentiation, using $\omega=k$, gives
$$
\frac{dU}{dt}=L^2E_0^2\cos(2\omega t).
$$
Thus
$$
\boxed{\int_SP\cdot dS=-\frac{dU}{dt}},
$$
confirming the integral identity when $J=0$.

Solved by gpt-5.6-sol high.