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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 ...


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 ...


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{...


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}...


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: $$z^2+{(\sqrt{ x^2+y^2}-R)}^2-r^2=0$$


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: $$ 2 \pi R^2/3 - R ( 1 - \cos (\pi/6) ) \cdot 2\pi R \cdot * 6/2 $$ $$ \approx 3.75782 $$...


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

Let $\Sigma_{g}^{1}$ be denote a surface of genus $g$ with one boundary circle. Let $\iota_g\colon \textrm{S}^{1} \to \partial\Sigma_{g}^{1}$ be a homeomorphism on the boundary. For convenience later on, we write $\textrm{S}^{1} = [0,1]/\{0 \sim 1\}$, and we fix base points $x_g = \iota_g(0) \in \Sigma_{g}^{1}$. I assume that the following statement is ...


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 ...


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 $$K = \frac{eg - f^2}{EG-F^2} = \frac{e\times 0 - 0^2}{EG-F^2} = 0.$$



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