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Apply the product-to-sum identity directly to the stated modulation:The carrier has angular frequency $\omega$ and the two modulation sidebands have angular frequencies $\omega \pm {\omega }_m$, each with amplitude $b/2$.

Propagation qualification. In the literal formula, all three terms have the same spatial wavenumber. That identity is correct, but a time-only amplitude applied everywhere is not generally a freely propagating wave at speed $c$. For $k=\omega /c$, applying ${\partial }_t^2-c^2{\partial }_x^2$ to the stated field gives $A_0^{\prime \prime }\cos (\omega t-kx)-2\omega A_0^{\prime }\sin (\omega t-kx)$, which at $x=t=0$ equals $-b{\omega }_m^2\ne 0$ when the modulation is nontrivial.

A physical modulation launched at a boundary and travelling without dispersion has envelope $a+b\cos [{\omega }_m(t-x/c)]$. The same expansion then has sideband wavenumbers $k_{\pm }=(\omega \pm {\omega }_m)/c$, while its value at $x=0$ agrees with the given time-modulated signal. This distinguishes the requested frequency decomposition from an unsupported fixed-wavenumber propagation claim.

The integration region is the triangle $0\le y\le x\le \pi$. Reversing its order givesThe factor of $x$ from the inner interval cancels the denominator. At $x=0$, $\sin x/x$ has a removable singularity, so no divergent integral is being hidden by this change of order. Fubini's theorem applies to its continuous extension on the bounded triangle.

A perpendicular geometric cross-section at position $x$ is a disk of radius $r=x^2$, so its area is $\pi x^4$. Slicing along the axis givesThe result uses the coordinate units implicit in the source's relation between radius and axial coordinate.

The gradient is $\nabla f=(2x+y,x)$ and at $(1,1)$ it is $(3,1)$. For a unit vector $\hat{\mathbf{n}}$, the directional derivative is $D_{\hat{\mathbf{n}}}f=(3,1)\cdot \hat{\mathbf{n}}$. The Cauchy-Schwarz inequality bounds its absolute value by $\sqrt{10}$, with equality in the parallel or antiparallel direction. ThusThe plus sign gives greatest increase; the minus sign gives greatest decrease, with the same magnitude. Stationary directions at the specified point have zero first derivative, so they are perpendicular to the gradient:These are tangents to the local level curve, not a claim that $(1,1)$ is itself a stationary point of the two-variable function.