New answers tagged

1

This is actually one of the standard formulas in a first-year calculus course for the surface area of a surface of revolution. You want to compute the integral $$\int_0^{2\pi}\int_0^r \phi(u)\sqrt{1+\phi'(u)^2}\,du\,d\theta.$$ (This is the surface obtained by rotating the graph $y=\phi(x)$ about the $x$-axis.) Alternatively, you can derive this using the ...


1

So after some time I got both of the algorithms working. In scenes with surfaces of lower degree, algorithm by Mann and DeRose seems to be faster, in other scenes (about 80% of scenes I test my ray tracer with) the algorithm by Sederberg is faster and requires a constant amount of memory so it is more suitable for GPU (no dynamic allocation in OpenCL 1.2). ...


1

Let's call $f_0(y)$ the values on the segment $\{0\}\times[0,1]$, $f_1(y)$ the values on $\{1\}\times [0,1]$, and similarly $g_0(x)$ and $g_1(x)$. A continuous function interpolating them is $$ \frac{[(1-x)f_0(y) + xf_1(y)]y(1-y) + [(1-y)g_0(x) + yg_1(x)]x(1-x)}{x(1-x) + y(1-y)} $$


1

Yes, $X \setminus U$ must be compact. To prove this, first let's enlarge the set $K$ somewhat: we may include $K$ into a subset $K \subset Y$ such that $Y$ is a smooth, compact, surface-with-boundary. This is not too hard to see: take a collection of smooth balls whose interiors cover $K$; by compactness finitely many of these balls suffices; we may perturb ...


0

Some simplifications. The matrix of the quadratic form part is half the Hessian matrix of second partial derivatives, so indeed $$ A = \begin{pmatrix} 0 & -2 & 1 \\ -2 & -3 & 2 \\ 1 & 2 & 0 \end{pmatrix}.$$ As it happens, $\det A \neq 0.$ As a result, there really is a "center", call it column vector $C,$ given by $$ 2 A C = \left( \...


0

It's a hyperbola of two sheets. You can determine this in two ways: 1) Examine the signs and powers of the variables (hyperbola of two sheets has two negative signs on two out of the three variables of degree two). 2) Examine the traces of the equation by setting one of the variables to a constant and then identify the two dimensional equation.


3

Your idea is good but note that the equation $$ -3y^2 - 4xy + 2xz + 4yz = 2x + 2z - 1 $$ does not define the intersection of the surface described by the quadratic form with the plane. Such an intersection would have been described by two equations and not a single equation: $$ -3y^2 - 4xy + 2xz + 4yz = a, \,\,\, 2x + 2z - 1 = b. $$ What works better ...


2

The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$ Setting $y=0$ gives $$\frac{x-\alpha}{l_1}=\frac{0-\beta}{m_1}=\frac{z-\gamma}{n_1}\implies x=-\beta\frac{l_1}{m_1}+\alpha,\quad z=-\beta\frac{n_1}{m_1}+\gamma\tag1$$ Also, $$\frac{l_1}{m_1}=\frac{x-\alpha}{y-\beta},\quad \frac{n_1}{m_1}=\frac{z-\gamma}...


0

The given plane of intersection has direction cosines $( \frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})$, surface normal direction along major diagonal of a unit cube. Ratio of true and projected areas is: $$ \dfrac{{\dfrac{\sqrt3 \cdot {\sqrt2}^2}{4}}} {{\dfrac{1^2}{2}}} = {\sqrt3}$$ So, the slant area of cut ellipse is $ {\sqrt3} \pi.$


0

You are intersecting the cylinder $x^2+y^2\leq1$ with the plane $x+y+z=1$. If $\theta$ is the angle between the $z$-axis and the normal of the plane then $$\cos\theta=(0,0,1)\cdot{1\over\sqrt{3}}(1,1,1)={1\over\sqrt{3}}\ .$$ If you project a piece of surface $S$ orthogonally onto the $(x,y)$-plane under such circumstances then the area $\omega(S)$ is ...


0

For a surface in 3D space, parametrization needs two parameters. (a curve needs one parameters.) You can use (x, y) as two parameters. The surface area is, $$S=\iint \limits_D \sqrt {f_x^2+f_y^2+1} dxdy$$ Here function $f$ is defined as, $$z=f(x,y)=1-x-y$$ and the domain is the circle $$x^2+y^2<1$$ This gives your the answer, $$\sqrt3 \pi$$


1

The standard formula for area of a spherical segment is $ 2\pi R *$ Axial length. Can be derived more easily as $ \int 2 \pi y \, ds = \int_{R-h}^R 2 \pi y \, \sqrt {1+y{\prime}^2} dx = 2 \pi R h $ For R =1,what is left after subtracting from hemi-sphere area is: $$ 2 \pi R^2/3 - R ( 1 - \cos (\pi/6) ) \cdot 2\pi R \cdot * 6/2 $$ $$ \approx 3.75782 $$...


3

The union of $S_{G}$ and its reflection across the $(x, y)$-plane is a unit sphere with six circular caps removed. Each cap is a zone cut by planes separated by $1 - \frac{\sqrt{3}}{2}$, and so has area $\pi(2 - \sqrt{3})$ by Archimedes' theorem. That is, the area of $S_{G}$ is $$ \tfrac{1}{2}\bigl[4\pi - 6\pi(2 - \sqrt{3})\bigr] = \pi\bigl[2 - 3(2 - \sqrt{...


1

Hint: By simmetry, the area of the surface of the half sphere which lies above the equilateral triangle $OP_6P_1$ is $\frac16S_G$. Now, $\triangle OP_6P_1$ is the set bounded by the $x$ axis, and the lines $y=\sqrt{3}x$ and $y=\sqrt{3}(1-x)$. In polar coordinates this region is $$\left\{(\theta,r)\in\mathbb{R}^2:0\leq\theta\leq\frac{\pi}{3},0\le r\le\frac{\...


1

The relative long exact sequence gives us the exactness of $H^1(M) \to H^1(\partial M) \to H^2(M,\partial M) \to H^2(M)$. The last term is zero ($M$ deformation retracts onto a 1-complex; to see this, think about the picture of closed manifolds as polygons mod edge identifications). Call $M'$ the manifold you get when you glue in discs to each of the ...


0

Only because $a'$ has constant length $1$ can you say you want to show $a''=0$. In general, the $v$-curves will be lines if and only if $X_{vv}$ is a multiple of $X_v$ at every point. OK, so here's what you need to do. Calculate the Gaussian curvature, using $E$ and $G$, and then compare this with the extrinsic formula for the Gaussian curvature (in terms ...


0

Parametrization of the surface cut out by one cylinder The image above shows the intersection of the hemisphere with the cylinder whose axis passes through $(0,1,0)$ and whose equation is $$ x^2+(y-1)^2=1 $$ The surface cut out by the cylinder can be parametrized with Cartesian coordinates as $$ \vec\Sigma=\left(x,y,\sqrt{4-x^2-y^2}\right),\quad 0\le x^...


1

We are talking here about the half-cylinder in $x$-direction, intersecting the $(y,z)$-plane in the semicircle $$\gamma:\qquad y=\cos\phi,\quad z=\sin\phi\qquad(0\leq\phi\leq\pi)\ .$$ Through each point $(y,z)\in \gamma$ there is an infinite stalk in $x$-direction, but only the part $${z\over a}\leq x\leq{z\over b}$$ of length $$\ell(\phi):={z\over b}-{z\...


1

There is a symmetry. There is as much area between the planes on one side of the line intersection of the planes ($x=z = 0$)as on the other side of the line. The total area then is double the area in 1/2 the cylinder. Lets rewrite the planes as: $x = z/a$ and $x = z/b$ covert to polar: Surface area = $\int_0^{\pi}\int_{(r/b) \sin\theta}^{(r/a) \sin\...


1

According to $z=ax$ and $z=bx$ and that $a>b>0$, you see that $z/a\leq x\leq z/b$. In fact, this shows the variation of $x$. You can easily find the variations of $y$ and $z$ using polar coordinates. Since $z\geq 0$, so we have $0\leq \theta \leq \pi$ and $0\leq R\leq r$ which $(R, \theta)$ are polar variables.


1

Some representations of the ellipsoid: 1. $$ (u - u_c)^\top A (u - u_c) = 1 $$ where $u = (x,y,z)^\top$ and $u_c = (x_c, y_c, z_c)^\top$ and $A$ is a definite matrix with eigen values $r_a^{-2}$, $r_b^{-2}$ and $r_c^{-2}$ 2. $$ u = u_c + u_x x + u_y y + u_z z $$ where $A = (u_x, u_y, u_z)$ is a regular $3\times 3$ matrix. Where one can choose $u_x$, $u_y$...


1

Note that $$X(t, s) = f(t) + sf'(t),$$ where $f(t) = (t, t^2, t^3)$ (The precise forms does not matter). Then $$X_t = f' + sf'' , \ \ X_s = f'.$$ This implies that $f''$ lies in the tangent plane of the surface. Since $X_{ss} = 0$, and $X_{st} = f''$ lie in the tangent plane, we have $$K = \frac{eg - f^2}{EG-F^2} = \frac{e\times 0 - 0^2}{EG-F^2} = 0.$$


0

No way the tangent curvature can be fixed .For example you can draw on a sphere a geodesic, any small circle which can be any displaced arbitrary parallel, all living on the sphere. You can have them constant when deforming together isometrically.In bending both $k_g, K$ (geodesic and Gauss curvatures) remain constant. But you cannot derive one from the ...


0

No, and no. Consider a torus with "core" radius $r$ and "tube radius" $s$, lying on a table like a donut. A point i=$P$ in contact with the table surface lies on a circle of radius $r$ and an orthogonal circle of radius $s$, both geodesic, so the Gaussian curvature at $P$ is $rs$. Now fix $k$ and pick $r = k/s$. Then $rs = k$, so for any such $s$, we ...


2

This surface is a torus with eccentric dimension at z=0 equalling R for tube centre and tube radius r. In the equation of the circle $$z^2 +(x-R)^2-r^2=0$$ you have replaced $ x $ by $ \sqrt{x^2+y^2} $ which is how we can rotate the circle around $z$ axis to result in a torus: $$z^2+{(\sqrt{ x^2+y^2}-R)}^2-r^2=0$$


0

I realize the formula I've reached was not as helpful as the initial one for F. This surface is a torus with outer radius R and inner radius r.


4

From the looks of the plot on the left, this seems to be what MATLAB calls the peaks() function: $$\begin{align*}\mathtt{peaks}(x)=&3(1-x)^2\exp\left(-x^2-(y+1)^2\right)-10 \left(\frac{x}{5}-x^3-y^5\right) \exp\left(-x^2-y^2\right)-\\&\frac13\exp\left(-(x+1)^2-y^2\right)\end{align*}$$ A look at the M-file for this function tells me that Cleve Moler ...


0

For volumes the "geometric center" usually refers to the medial axis; for algorithms to compute it see for instance this paper. However note that the medial axis of 3D volume can be topologically complex (consisting of sheets and curves glued together) and you will require significant post-processing to get anything resembling a developable surface from the ...



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