Use the same axes as in (a) and choose the outward orientation. On the curved spherical lune, $\mathbf{n}={\mathbf{e}}_r$ and $dA=a^2\sin \theta d\theta d\phi$. Its vector area isThe $y$ and $z$ components cancel. The $x$ component is $a^2(\pi /2)2\sin (\beta /2)$, henceHere ${\mathbf{e}}_x$ points along the azimuthal bisector; another coordinate choice rotates this vector accordingly. Its magnitude is the signed projected area, not the scalar curved area $2\beta a^2$. Reversing the orientation reverses its sign.

Orient the common axis of the wedge faces along $z$ and its azimuthal bisector along positive $x$. Reflection symmetries give $\bar{y}=\bar{z}=0$. In spherical coordinates, use $0\le r\le a$, $0\le \theta \le \pi$, and $-\beta /2\le \phi \le \beta /2$. The volume isSince $x=r\sin \theta \cos \phi$, the first moment isDividing by the volume locates the centre of mass:This is the centroid formula for a spherical wedge.

For the sketch, $f$ is positive and decreases from ${\lim }_{\beta \to 0^{+}}f=3\pi /16$ to ${\lim }_{\beta \to 2{\pi }^{-}}f=0$, with $f(\pi )=3/8$. Indeed, on putting $q=\beta /2$, the sign of $f^{\prime }$ is the sign of $q\cos q-\sin q$. The negative of this expression has derivative $q\sin q>0$ for $0<q<\pi$ and starts at zero, proving strict decrease.

Equating the three Cartesian components givesThe first and third equations yield $\lambda +\mu =-2$ and $\lambda -\mu =4$, hence $\lambda =1$, $\mu =-3$. They also satisfy the middle equation, so the intersection is

The number of tails has a binomial distribution. The complement of at least two tails consists of zero tails or exactly one tail. Therefore

Treat the fair coin tosses as independent. Multiplying the ten head probabilities gives

The normal line to a plane curve is perpendicular to the tangent line, so its slope is the negative reciprocal $1/\sqrt{6}$. The required equation isThe normal direction also agrees with the gradient $(4x,2y)$ of the defining level function.

The positive horizontal coordinate is $x=\sqrt{3/2}=\sqrt{6}/2$. By implicit differentiation,At the selected point this gives $dy/dx=-\sqrt{6}$.

The usual interior condition for the binomial series is $|3x|<1$, giving $-1/3<x<1/3$ as its open interval of convergence. In this particular square-root case the full series also converges absolutely at both endpoints, so the maximal real range on which its sum equals the original function isFor an endpoint check, write $\sqrt{1-u}=1-{\sum }_{n\ge 1}{b}_n{u}^n$, where $b_n=a_{n-1}-a_n>0$ and $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\to 0$. Then $\sum b_n=1$ by telescoping, establishing absolute convergence at $u=\pm 1$. The square-root binomial series at its endpoints proves continuity of the sum there. The endpoint values are $\sqrt{2}$ at $x=-1/3$ and $0$ at $x=1/3$. Outside this range the power-series terms do not tend to zero. A three-term truncation remains an approximation, rather than an exact formula throughout the interval.

Use the binomial series with exponent $1/2$ and argument $-3x$:Thus the first three nonzero terms are

For this parametric curve, the chain rule givesThe quotient is a finite slope when $\sin t\ne 0$. At $t=0$ and $t=\pi$, the horizontal component of the tangent vanishes while the vertical component does not; these have vertical tangent lines, rather than missing finite values of the displayed quotient.

The maximum of sine is one. The allowed parameter values giving $\sin (3t)=1$ areSubstituting each into $x=\cos t$ givesAll three parameter values lie in the original interval, so none of these maxima is discarded.

Within the requested interval, the function is symmetric about $x=\pi /2$ because $\sin (\pi -x)=\sin x$. Its derivative $\cot x$ is positive before $\pi /2$, zero there, and negative afterwards. The negative second derivative shows that this is a strictly concave function. Consequently its unique maximum is $(\pi /2,0)$. It tends to $-\infty$ at both ends, giving vertical asymptotes $x=0$ and $x=\pi$.

On any interval where $\sin x>0$, the chain rule gives $y^{\prime }=\cos x/\sin x=\cot x$. Differentiating again,The domain condition is necessary because the real natural logarithm requires a positive argument.

For an oriented surface $S^{\prime }$ with compatible boundary direction, the Stokes theorem statesLet $H$ denote the removed patch and $S^{\prime }=S\setminus H$ the remaining cylinder. Keep the outward orientation inherited from the closed cylinder. Everywhere on the removed curved patch, the normal curl component is $-\alpha R$, so its flux is $-\alpha RA$, where $A$ is its actual curved area. Part (d) then givesConsequently the circulation around a hole in an oriented surface isThe positive direction here is induced by the remaining outward-oriented surface; locally it runs clockwise around the hole when viewed from outside. If the edge is instead traversed positively around the removed patch, the direction reverses and the answer is $-\alpha RA$. The question does not specify an edge direction, so the sign must be accompanied by this convention. The result depends only on the removed curved area, not on its shape. If the patch has several boundary components, use the sum of their induced line integrals. Since $[\alpha ]=1/(\text{length}\cdot \text{time})$, the result has the units of a velocity line integral, length squared per time.

Separate the closed cylinder into its curved side and its two end disks, all with outward normal vectors. On the side, $\rho =R$ and $\mathbf{n}=\hat{\mathbf{\rho }}$, soAt the top, $z=h/2$ and $\mathbf{n}=\mathbf{k}$, giving $(\nabla \times \mathbf{v})\cdot \mathbf{n}=\alpha h$. At the bottom, both $z$ and the axial normal vector change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi \alpha R^{2}h$. Adding all three surface integrals,This explicit cancellation is consistent with the fact that the divergence of a curl is zero. The two disk contributions add rather than cancel.

The curl gives the vorticity:Since $x\mathbf{i}+y\mathbf{j}=\rho \hat{\mathbf{\rho }}$, the requested form isAt the axis the radial unit vector is undefined, but its coefficient vanishes and the Cartesian expression remains smooth. The radial vorticity reflects the change in rotation rate with height.

For a sufficiently smooth vector field on a volume $V$ and its closed boundary $S$, the divergence theorem stateswith the unit normal directed outwards. HereHence the net volume flux through every closed surface in the liquid is zero. This verifies incompressible flow directly. It does not require the velocity to vanish pointwise on an arbitrary closed surface; inward and outward contributions can cancel.