# Tag Info

13

You have asked a good number of questions. I'll answer the one in the title. The point is that differential forms are "the things you can integrate on manifolds". Manifolds are more general objects than open subsets of $\Bbb R^n$, and in some sense one of the reasons one wants to introduce forms. Suppose you have a 1-form $\alpha$ on a manifold $M$, and a ...

4

You tag this as multivariable-calculus, implying you want a double or triple integral, but this can be done with a single integral. Your desired solid is the rotation of the upper half of your diagram about the $x$-axis. And that is double the rotation of the upper-right quarter of your diagram about the $x$-axis. The lower bound of $y$ of that upper-right ...

4

On $y=x^2$ you have $$\lim\limits_{(x,y)\to(0,0)}\left|\frac{y}{x^2}\right|\exp\left(-\left|\frac{y}{x^2}\right|\right) =\lim\limits_{(x,y)\to(0,0)}\exp(-1)=\frac{1}{e}$$ Denote $$f(x,y)=\left|\frac{y}{x^2}\right|\exp(-\left|\frac{y}{x^2}\right|)$$ on $y=0$ we have $f(x,0)=0$ and therefore $$\lim\limits_{(x,y)\to(0,0)}f(x,y)= ... 3 Let (x_1, y_1), (x_2, y_2) be two points in \mathbb R^2 so that y_1 < y_2 and$$x_1 e^{y_1} = x_2 e^{y_2} =: C.$$Consider$$h(t) = f\left(Ce^{-t}, t\right), t\in [y_1, y_2] .$$Then$$h'(t) = -f_x(Ce^{-t}, t) Ce^{-t} + f_y( Ce^{-t}, t) = 0$$as x f_x =f_y . Thus h(t) is a constant function and so$$\tag{1} f\left(x_1, y_1\right) = f(x_2, ...

3

Hint: $\log(x+e^y)$ is differentiable at $(0,0)$ with gradient $(1,1)$, so $$\log(x+e^y) = x+y+o(x,y)$$

3

If $f$ is constant, any $g$ will do. Otherwise there is a point $(p,q)\in{\mathbb R}^2$ with $\nabla f(p,q)\ne0$. Let $f(p,q)=:c$, and assume that $f_y(p,q)\ne0$. By the implicit function theorem there is a $C^1$-function $$\psi:\quad x\mapsto y:=\psi(x)\ ,$$ defined in some neighborhood $U=\ ]p-h, p+h[\$ of $p$, with $\psi(p)=q$, and ...

3

Just compute the second derivative, we have, by the chain rule $$\def\norm#1{\left|#1\right|} Df(x)h = \frac 1{(1 + \norm{x}^2)^{1/2}} \cdot \def\<#1>{\left<#1\right>}\<x, h>$$ Hence, $$D^2f(x)[h,k] = -\frac 1{(1 + \norm x^2)^{3/2}}\<x,h>\<x,k> + \frac{1}{(1 + \norm x^2)^{1/2}}\<h,k>$$ So, we have \begin{align*} D^2 ...

2

First, parametrize the surface $S$: $$x=x, \quad y=y, \quad z=\sqrt{x^2+y^2},$$ with $x,y \in D := \{(x,y)\;|\; x^2+y^2\le h^2\}$. Second, compute $\vec{r}_x \times \vec{r}_y= (\frac{-x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},1)$. Third, note that you can rewrite your integral as follows: $$\iint_S \vec{F}\cdot d\vec{S},$$ with ...

2

If you let $\vec n=(a,b,c)$ then the integrand function can be written as $f(ax+by+cz)=f(\vec n\cdot\vec r)$. Choose the reference frame so that $\vec n$ lies along the $z$ axis: it follows that $f(\vec n\cdot\vec r)=f(nz)$, where $n=\sqrt{a^2+b^2+c^2}$. Integrate then in spherical coordinates: $$\int \int_F f(ax+by+cz)dS = \int_{0}^{2\pi}d\phi\int_0^\pi ... 2 The function xy is the height at each point, so you have bounded z between 0 and xy quite naturally, by integrating the height of your object over the xy-plane. The more thurough way of doing it would, of course, be to calculate the triple integral$$ \int_0^1 \int_0^x\int_0^{xy} 1\, dz\,dy\,dx $$but as you might see, the innermost integral ... 2 The inverse function theorem tells you that for each x\in A, there is an open neighborhood U of x and an open neighborhood W of f(x) contained in f(U) such that f^{-1}:W\to U is differentiable. So f(A), being the union of these open neighborhoods W around each point f(x)\in f(A), is open. Since differentiability is a local property, it ... 2 To calculate the surface integral over a surface we must start by finding the surface normal to the surface at each point, {\rm d}{\bf S}(x,y,z) = {\bf A}(x,y,z){\rm d}x{\rm d}y + {\bf B}(x,y,z){\rm d}x{\rm d}y + {\bf C}(x,y,z){\rm d}y{\rm d}z. With this in hand we simply take the dot product with the vector field {\bf V}, derive the correct integration ... 2 Yes, you are correct. Known the normal vector v at the a point P=(x_P,y_P,z_P) of the surface this gives you the equation of the tangent plane in P :$$v \cdot (x-x_P,y-y_P,z-z_P)=0.$$2 Just consider the directional derivatives:$$\lim_{h\to0, h>0} \frac{f(h,0) - f(0,0)}{h} = \frac{|h|}{h} = \frac{h}{h} = 1$$and$$\lim_{h\to0, h<0} \frac{f(h,0) - f(0,0)}{h} = \frac{|h|}{h} = \frac{-h}{h} = -.1$$This means even the partial derivatives wrt. x do not exist. 2 Let g be the restriction of \Phi to the boundary of the square. Your boundary data determines g up to a constant. You can then take this g and solve the corresponding Dirichlet problem; standard theory tells that solutions exist uniquely. Therefore you can conclude that solutions are unique up to shifting by constants. (The fact that shifting a ... 2 Don't do it by definition. Do it by using theorems you know: The product of two continuous functions is continuous The composition of two continuous functions is continuous The sum and difference of two continous functions is continuous The functions f(x,y)=x and g(x,y) = y are continuous. You only need to be a little careful about when \frac fg is ... 2 just out of curiosity, we know that x^2/a^2+y^2/b^2=1, then by AM–GM inequality$$1=\frac{x^2}{a^2}+\frac{y^2}{2b^2}+\frac{y^2}{2b^2}\geq 3\sqrt[3]{\frac{x^2}{a^2}\cdot\frac{y^2}{2b^2}\cdot\frac{y^2}{2b^2}}$$the equality holds when$$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$2 Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for g_y, we have$$\begin{align} y^2 &= 2b^2\lambda x \\ 2xy &= 2a^2\lambda y \end{align}$$which upon eliminating \lambda gives$$ a^2y^3 = 2b^2x^2y, $$so either y=0 or a^2y^2=2b^2x^2. Substituting this into the constraint: ... 2 Since you're applying the exponential function to a vector argument, you must be doing it element-wise. The differential of a scalar function f(s) is a scalar given by$$df = f' ds$$Similarly, when applied element-wise to a matrix argument (X) the differential is a matrix given by$$df = f'\circ dX$$where \circ represents the Hadamard product. Now let ... 2 (I'll keep it simple, so this explanation sacrifices a lot of generality.) There are two ways to apply a linear transformation to a vector field: multiplication and conjugation. What you've observed is the difference between the two. Multiplying a vector field (on the left) rotates all the arrows in-place. If you start with a vector field that looks like ... 2 To visualize the transformation of a vector field you first have to visualize the vector field. In your case, you are rotating the identity field in \mathbb R^2. Consider an arbitrary point in \mathbb R^2, say (3,4). The identity vector field assigns the vector 3\hat i + 4\hat j to this point. You imagine an arrow of length 5 being drawn with its ... 2 Consider a small piece of volume \Delta V at position (x,y,z). There exist a mirror piece at (-x,y,z) with equal volume. The contribution of these two mass elements to the integral \int_V x\sigma{\rm d}V is (see picture below)$$x \sigma( \rho) \Delta V + (-x)\sigma( \rho) \Delta V = 0$$since both of these elements have the same distance \rho ... 2 You have almost answered your own question... When switching to polar coordinates, we do not need \theta\rightarrow0, in fact we must allow \theta to do anything, as long as r\rightarrow0. This shows in fact that the function cannot be made continuous, since the limit will depend on \theta. HINT an even easier approach is to consider the limit ... 2 First of all, all norms on \Bbb R^{n} are equivalent (do you know what that means?). As a consequence, if a function is continuous with respect to one norm, it is continuous with respect to every norm, so you can just pick the norm that makes the proof of continuity easiest based on the given function. Also, just as we prove continuity in the ... 1 Already, f is a continuous function from \mathbb{R}^2 \setminus \{(0,0)\} to \mathbb{R}, and one might ask whether we can extend the domain of f to all of \mathbb{R}^2 (that is, define f(0,0)) in such a way that f is a continuous function from \mathbb{R}^2 to \mathbb{R}. Suppose f could be extended in some way. Then (a general ... 1 Hint: try and factorize x^4-y^4. 1 Your function F is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc. Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows:$$ F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad ...

1

As an alternative approach, if you happen to know some handy formulas: The volume in $S_2$ that is outside $S_1$ and $S_3$ is equal to the volume of $S_2$ (which is $\frac{4\pi}{3}$) less the volume of four endcaps, each of "height" $1/2$. The expression for a volume of a spherical cap of height $h$ in a unit sphere is  V_\text{cap} = \frac{\pi ...

1

Your computations appear correct with the exception that $(0,0)$ (the origin) is now just represented by $r=0$; you will also need to state that due to the limit being independent of $r$, you may approach from different angles and get different limits (e.g., approaching $r=0$ with $\theta=0$ gives a limit different than when $\theta=\pi/2$). This will prove ...

1

You could think of $\sigma$ as density of the sphere, and since the only variable in $\sigma = k \left(1 - (\rho / a)^3 \right)$ is $\rho$, the distance from a point to the origin, you can see that for any spherical shell inside the sphere the density is uniform(only for a spherical shell). From the definition of centre of mass, your integrals represent the ...

Only top voted, non community-wiki answers of a minimum length are eligible