Use the spherical polar coordinates volume element $dV=r^2\sin \theta drd\theta d\varphi$. Since $1+\cos \theta \ge 0$ throughout $0\le \theta \le \pi$, the solid has radial range $0\le r\le 1+\cos \theta$ and full azimuthal range $0\le \varphi <2\pi$. Its volume isPut $u=1+\cos \theta$, so $du=-\sin \theta d\theta$ and the endpoints become $2$ and $0$. ConsequentlyThe polar equation describes the entire three-dimensional surface, so the spherical polar coordinates volume element, rather than a two-dimensional polar area element, is required.

The region of integration is $0\le y\le 1$, $y\le x\le 2-y$. It is the closed triangle with vertices $(0,0)$, $(2,0)$, and $(1,1)$, bounded by $y=0$, $x=y$, and $x=2-y$. Each horizontal slice runs from the rising left edge to the falling right edge.

Integrate with respect to $x$ first:The triangle has area one, but the nonconstant integrand makes the double integral a weighted area.

Here the vertical bars denote determinants. Since $B=B^{-1}$, taking determinants gives $(\det B{)}^2=1$. The condition $\det B>0$ selects $\det B=1$. By the multiplicative property of the determinant,The hypotheses determine this determinant, while allowing many possible matrices $A$ and $B$.

The homogeneous linear system has coefficient matrixNonzero solutions exist exactly when its determinant vanishes. Expanding along the first row givesThus $\lambda =1$ or $-6$. The first equation always gives $x=3z$. For $\lambda =1$, the second then gives $y=-2z$; the third imposes the same relation. For $\lambda =-6$, the second gives $y=3z/2$, again consistent with the third. Therefore the nonzero solutions areFor every other $\lambda$, the matrix is invertible and only the zero solution exists.

Maximize the total surface area $A=2\pi r(r+h)$ with the constraint $g=r^2+h^2/4=1$. The Lagrange multiplier equations areFrom the second, $\lambda =4\pi r/h$. Substitute into the first and multiply by $h/(2\pi )$ to obtain $2rh+h^2=4r^2$. Thus the aspect ratio $q=h/r$ satisfies $q^2+2q-4=0$. Since $q>0$,This is the maximum-area cylinder inscribed in a sphere. The dimensions and area, although not needed to specify the ratio, areThere is only one positive stationary ratio. At the degenerate boundaries $r=0$ and $h=0$, the limiting areas are zero and $2\pi$, both below this value, confirming the maximum.

Use a Lagrange multiplier to maximize $V=\pi r^{2}h$ subject to $2\pi r(r+h)=A$, with $r,h>0$. The equations $\nabla V=\lambda \nabla A$ areThe second equation gives $\lambda =r/2$. Substituting into the first gives $2rh=r(2r+h)$, hence $h=2r$. The constraint then gives $A=6\pi r^2$, soThis is the same aspect ratio as the minimum-area closed cylinder at fixed volume. To confirm maximality directly, eliminate $h$: $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi )}$. Its only stationary point is the value above, its second derivative there is negative, and the volume tends to zero at either end of the interval.

Compute the curl in Cartesian coordinates:The three components are $6y{z}^2-6y{z}^2$, $0-0$, and $e^x-e^x$. Consequently

For $R=1$ and $z_1=0$, the shortest path on a circular cylinder has lengthIts graph is triangular, with endpoint values zero and maximum $\pi$ at ${\varphi }_1=\pi$. In particular, increasing the angle to a full turn brings the endpoint back to the starting point.

For $R=1$ and $z_1=1$, the corresponding length isThis graph has endpoint values one, a maximum $\sqrt{1+{\pi }^2}$ at $\pi$, and symmetry about $\pi$. Each half is smooth, with horizontal slope at the outer endpoint; at $\pi$ the two halves meet at a corner, because the minimizing winding direction changes.

On the circular cylinder, $\rho =R$ is constant, so the given line element reduces to $d{s}^2=R^{2}d{\varphi }^2+d{z}^2$. Unroll the circular cylinder with $u=R\varphi$. The line element is then $d{s}^2=d{u}^2+d{z}^2$, so the usual Euclidean geometry formula measures arc length in the $(u,z)$ plane.

The endpoint has infinitely many lifts $(R({\varphi }_1+2\pi k),z_1)$, with $k\in \mathbb{Z}$. For each lift, the straight line segment is shortest; its length is $\sqrt{R^2({\varphi }_1+2\pi k{)}^2+z_1^2}$. Comparing the lifts gives the unrestricted shortest path on a circular cylinder:If $\delta$ is a minimizing angular increment, a shortest geodesic is parametrized by $(\rho ,\varphi ,z)=(R,\delta t,z_{1}t)$, $0\le t\le 1$. This is a helix when both increments are nonzero, and a circular arc or vertical generator in the limiting cases. A nonminimal lift also gives a geodesic, but it is only shortest within its fixed winding class.

For $0\le {\varphi }_1\le 2\pi$, put $m=\min ({\varphi }_1,2\pi -{\varphi }_1)$. Then $\boxed{L_{\min }=\sqrt{R^2{m}^2+z_1^2}}$. At ${\varphi }_1=\pi$, both winding directions give distinct shortest paths of equal length.

The conversion from cylindrical coordinates to Cartesian coordinates isHere $\rho \ge 0$ is the distance from the $z$ axis and $\varphi$ is measured around that axis. A full turn $\varphi \mapsto \varphi +2\pi$ describes the same point.

The summands $u_n=n^4/2^n$ are positive. Apply the ratio test:ThereforeFor a direct comparison interpretation, the ratio is eventually at most any fixed number between $1/2$ and $1$, such as $3/4$. The tail is then bounded by a convergent geometric series. Thus the exponential denominator wins over the fourth-power numerator strongly enough to make the whole series summable.