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The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

For the parametrized surface in i,These tangent vectors are orthogonal, and their Euclidean norms are $\sqrt{(c-a{)}^2+(d-b{)}^2}$ and $\rho$, respectively. Their cross product therefore has Euclidean norm $\rho \sqrt{(c-a{)}^2+(d-b{)}^2}$. Integration givesThis is the lateral surface area of a conical frustum, with no end disks included.

Let $B=A^{T}A$. It is a symmetric matrix, and ${\mathbf{v}}^{T}B\mathbf{v}=\Vert A\mathbf{v}{\Vert }^2\ge 0$. By the spectral theorem for real symmetric matrices, choose an orthonormal basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ of eigenvectors of $B$, with nonnegative eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

For each $i\ne j$, take the unit parallelogram parametrized by $s{\mathbf{v}}_i+t{\mathbf{v}}_j$, $0\le s,t\le 1$. Its surface area is $1$. Moreover,The image parallelogram consequently has area $\sqrt{{\lambda }_i{\lambda }_j}$. By the assumption this equals $1$ for all three pairs, givingNo eigenvalue can be zero. Dividing the equations and using positivity yields ${\lambda }_1={\lambda }_2={\lambda }_3=1$. HenceThus testing all surfaces is stronger than necessary: the three unit parallelograms in these principal directions already determine the result. This is the three-dimensional characterization of linear surface-area isometries.

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

A parametrization of the cylindrical patch isThe last range uses the planes $x=0$ and $x=y$. The surface area element is $|{\mathbf{r}}_x\times {\mathbf{r}}_{\theta }|dxd\theta =2dxd\theta$. HenceIncluding or excluding the boundary curves does not change the surface area.

One sufficient version of the divergence theorem is the following. Let $R$ be a bounded region in ${\mathbb{R}}^3$ with a closed, piecewise smooth surface as its boundary, and let $\mathbf{F}$ be a continuously differentiable function with values in ${\mathbb{R}}^3$ on an open neighbourhood of the closure of $R$. With $\mathbf{n}$ the outward unit normal vector,The boundary surface integral is the total outward flux integral. If the boundary has several components, each has the outward orientation relative to the material of $R$; inner components therefore point into the holes. Edges and isolated vertices may be omitted when assigning the unit normal vector, since they have zero surface area.

Use cylindrical coordinates $(\rho ,\theta ,z)$. The right circular cone is described by $0\le z\le 1$ and $0\le \rho \le a(1-z)$, and the divergence of $\mathbf{F}=(0,0,z^2)$ is $2z$. Direct integration gives

To check the boundary flux integral independently, first consider the base $z=0$. Its outward unit normal vector is $-\mathbf{k}$, but $\mathbf{F}=0$ there, so its flux is zero. Parametrize the lateral surface byThe cross productpoints outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral isAdding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

The rotation axis is $L=\{(x,1,0):x\in \mathbb{R}\}$. The distance of the generating function graph from this axis is $\sqrt{x}$, so a parametrization isThe surface of revolution therefore has equation $(y-1{)}^2+z^2=x$: it is a circular elliptic paraboloid opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the surface area formula, soThe cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

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.

In cylindrical coordinates the spherical cap has $1<z<\sqrt{2}$, $0<r<\sqrt{2-z^2}$ and $0<\theta <2\pi$. Its mass density is $kz$ and the volume element is $rdrd\theta dz$. ThereforeThis is the mass of a spherical cap with height-proportional density with $R=\sqrt{2}$ and $c=1$.

For its curved surface area, write $z=\sqrt{2-r^2}$ with $0\le r\le 1$. The surface integral of one becomesIt agrees with the curved area of a spherical cap, $2\pi Rh$, for $h=\sqrt{2}-1$. The planar closing disk is not part of the requested curved surface.

The base is the disk $D:(x-a{)}^2+y^2\le a^2$, on which $x\ge 0$. The vertical thickness is $(p-q)x$. Using the symmetry of the disk about $x=a$, or integrating with $x=a+r\cos \varphi$, givesFor the curved cylindrical surface use $x=a(1+\cos \varphi )$, $y=a\sin \varphi$, with $0\le \varphi \le 2\pi$ and $qx\le z\le px$. The area element is $ad\varphi dz$. Its surface area is thereforeThe sloping planar faces are excluded from this curved surface area.

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently

The curved surface area excludes the two end disks, so $A=2\pi rx=2\pi \tan \alpha x(h-x)$. Completing the square givesThusIt differs from the volume optimum because the objective contains one power of the radius rather than two.

The scalar integrand still equals $10e$ throughout the unit sphere, and the upper hemisphere has the same surface area $2\pi$. Excluding its equator changes no area. ThereforeThere is no sign reversal between hemispheres because this scalar surface integral has no choice of normal orientation.

Writing $r^2=x^2+y^2+z^2$, the gradient is $\mathbf{v}=2e^{r^2}(x,y,z)$. Its divergence, equivalently the Laplacian of $\varphi$, isOn the unit sphere this scalar is the constant $10e$. The lower hemisphere has surface area $2\pi$, soThis is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

Take positive dimensions measured in metres. The volume is $V=lwh$ and the external surface area constraint is $S=lw+2lh+2wh=12$. There is one base and four walls, with no lid. The Lagrange multiplier equations for $V-\lambda (S-12)$ areThe first equation implies $\lambda >0$. Multiply the first by $l$ and the second by $w$ and subtract; this gives $2\lambda h(l-w)=0$, hence $l=w=s$. The last equation gives $\lambda =s/4$. Substituting into the first yields $sh=(s/4)(s+2h)$ and therefore $s=2h$.

The constraint now reads $s^2+4sh=3s^2=12$. ThusFor a global check, apply the arithmetic-geometric mean inequality to $lw$, $2lh$ and $2wh$, whose sum is $12$. Their product is $4V^2\le 4^3=64$, so no feasible box has $V>4$. Equality holds exactly at the dimensions found above. This is the maximum-volume open rectangular box at fixed surface area.

For $h>0$, parametrize the curved surface by $(r\cos \theta ,r\sin \theta ,r^2/a)$, with $0\le r\le \sqrt{ah}$ and $0\le \theta \le 2\pi$. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:Integrating givesThis is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The volume and centroid of a parabolic cup use the same geometry.

Use the embedded surface parametrizationIts tangent vectors have cross productwhose magnitude is $r\sqrt{1+4a^2{r}^2/b^4}$. Thus the surface area isWith $u=1+4a^2{r}^2/b^4$, the radial integral becomes $b^4(8a^2{)}^{-1}\int u^{1/2}du$. Evaluating its endpoints givesThis is the scalar-area part of the paraboloid sector surface measures.