# Tag Info

2

Your second parametrization does not trace out the surface of a cylinder, but instead gives a disk of radius $R$ in the $z=2z_0$ plane, centered on the $z$-axis.

0

Intuition: a tangent plane intersects a convex surface (e.g. a paraboloid) in exactly one point. If that's true, and substituting one equation into the other, we seek the only solution $(x,y)$ which satisfies $x^2+x+(ay^2+y+1)=0$. Applying the formula for the roots of a quadratic equation (twice) and demanding a single root, I got the condition for ...

0

The surface integral is given by: $$\iint \left\| {\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y} \right\| dx\, dy$$ Now when you transfer to cylindrical coordinates, you must convert your derivatives and add the Jacobian, so you must convert: $${\partial \mathbf{r} \over \partial x}={\partial \mathbf{r} \over \partial ... 0 Basically you can construct the surface of genus g, by gluing a disk \mathbb{D}^2 to a wedge of 2g circles, \bigvee\limits_{i=1}^{2g} \mathbb{S}_i^1. First we think the border of \mathbb{D}^2 as 2g-sided polygon, and we put labels in each side a_1,\ b_1,\ a_1,\ldots,\ a_g,\ b_g, and arrows as in the picture: Then we label each copy ... 0 Another way to find this is integrating on the lengths of the intersections of the surface of interest and the planes of the form y=k for -1\leq k \leq 1. I haven't tried it in a while but here it goes. The sections look like parenthesis joined by flat top and bottom. By parameterizing one of the cylinders with \theta; the length of the part that looks ... 0 The surface area is$$S=2\int\int_D \sqrt{1+f_x^2+f_y^2}dxdy$$where z=f=\sqrt{1-y^2} and so f_y=\frac{-y}{\sqrt{1-y^2}} so$$S=2\int\int_D \sqrt{1+\frac{y^2}{1-y^2}}dxdy=2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dxdy \\=2\int_{-1}^12dy=8 $$1 If you are looking at a hypersurface which is defined by an equation of the form F(x)=0, then you are looking at the set of point along which F does not change. The gradient of F points into the direction in which F changes most, so it is normal. In formulas: if c=c(t) is a curve in F=0, then F\circ c(t)=0, hence also it's derivative: ... 2 The principle curvatures are extrinsic quantities, the depend on the embedding of the manifold into some exterior manifold, and describe how the embedded manifold curves in that space. The curvature tensor of a Riemannian (or Lorentzian) manifold is an intrinsic quantity which measures to which extend covariant derivatives commute. 0 The surface X(t,\theta) is regular if X_t,X_\theta (the partial derivatives wrt each variable) are linearly independent for all (t,\theta)\in domain. Compute X_t\times X_\theta. Showing that it is nowhere zero would imply regularity. Let$$f(t)=\pm (arctan{(\sqrt{1-\mathbb{e}^{2t}})}-\sqrt{1-\mathbb{e}^{2t}}).$$Then$$f'(t)=\pm\frac{e^{2 t} ...

0


2

Here's another way. This produces a slightly rougher looking plot than Acer's, but it is much more intuitive to encode (just one line): plot3d(piecewise(abs(x)+abs(y) <= 1, (1+x^2)/(1+y^2), undefined), x= -1..1, y= -1..1);

1

Here is one way... Maple can solve that inequality in x and y which determines the domain. solve(abs(x)+abs(y)<=1); {0 <= x, 0 <= y, x <= 1, y <= 1 - x}, {0 <= x, x - 1 <= y, x < 1, y < 0}, {-1 <= x, 0 <= y, y <= 1 + x, x < 0}, {-1 - x <= y, -1 < x, x < 0, y < 0} From there we can (for this ...

Top 50 recent answers are included