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In Cartesian coordinates, the divergence uses the matching coordinate for each component:Each of these three partial derivatives vanishes.

Perpendicular vectors have zero dot product. Here $\mathbf{u}\cdot \mathbf{v}=3\cdot 4-2c$, so $12-2c=0$. Therefore

For a real-valued expression, $\ln x>0$ requires $x>1$. Applying the chain rule twice gives

The integrand is a continuous function, so the fundamental theorem of calculus differentiates the upper limit directly:No explicit antiderivative is needed.

For negative real $x$, the angle $u=\pi e^x$ lies strictly between $0$ and $\pi$. On that interval, $\sin u=1/\sqrt{2}$ has exactly the two solutions $u=\pi /4$ and $u=3\pi /4$. Dividing by $\pi$ and applying the natural logarithm yieldsBoth values are negative, and the restriction $0<u<\pi$ excludes all further periodic sine branches.

Put $u=\sin \theta$ and move all terms to one side. The polynomial factors asThe root $u=1$ yields the already supplied $\theta =\pi /2$. In the stated interval, $u=1/2$ gives $\pi /6,5\pi /6$, while $u=-1/2$ gives $-\pi /6,-5\pi /6$. Thus all the other solutions areThere are no endpoint solutions because $\sin (\pm \pi )=0$ is not one of the three roots.

Let the equilateral triangle have side length $a$. Its altitude is $H=\sqrt{3}a/2$, and its incircle has radius $r=H/3=a/(2\sqrt{3})$: the centre lies on a median, one third of the altitude above the base. The area of the triangle is $aH/2=\sqrt{3}{a}^2/4$, while the circle has area $\pi r^2=\pi a^2/12$. ThereforeThe side length cancels, as required for a dimensionless ratio between similar configurations.