# Tag Info

## New answers tagged integration

1

All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel). If ...

5

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...

3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$v = \frac{dx}{dt} = A x^2$$ You can regroup terms on the same variable: $$\frac{dx}{x^2} = A dt$$ And then do the integration: $$\int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...

13

It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 3 The general formula is indeed a double integral, so the most technically correct way to write it is $$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$ But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by ... 5 It is just a more compact notation. It is implied by the integration element$dA$that you are integrating over the surface. 1 The factor comes from the moment of inertia of the infinitesimal piece. In the disc method, each piece is a filled flat circle (a disc) of radius$r$, and the moment of inertia of a flat circle is$\frac{1}{2}mr^2$. The$\frac{1}{2}$accounts for the fact that the mass of the circle is distributed between the center and the edge. But in the shell method, ... 3 Maxwell's equations in curved spacetime are written in the form $$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with$F$the Faraday two-form,$J^a$the current four-vector,$\nabla$the covariant derivative and$[]$denotes antisymmetrization of the indices. In terms of exterior calculus they become: $$... 1 So on in three-dimensional Euclidean space we have an isomorphism between vectors and 1-forms, the usual way$$\eta_\mu = g_{\mu\nu} \eta^\mu.$$We also have an isomorphism between 1-forms and 2-forms, given by \star : dz\mapsto dx\wedge dy and cyclically. This isomorphism has a fancy name, the Hodge dual, if you want to know about it in general. Then if ... 1 Without specifying any particular scenario, and ignoring any proportionality constants, simply consider some general differential form \omega, and let this represent the electric flux through a closed surface which bounds some volume V. In classical electromagnetism, the Gauss law tells us that the flux through a closed surface is proportional to the ... 0 I will give you the general strategy for a problem of this kind. You know that acceleration is the instantaneous rate of change of position:$$a=\frac{dv}{dt}$$Through change of variables, this can also be equal to:$$a=\frac{dv}{dx}\times\frac{dx}{dt}.$$By chain rule:$$a=v\frac{dv}{dx}$$Knowing the acceleration as a function of position means you have: ... 3 The integral of a scalar function / vector field along a curve is defined with reference to a parametrization, e.g for a scalar$$ \int_C f \mathrm{d}s := \int_a^bf(r(t))\lvert r'(t)\rvert \mathrm{d}t$$where r: (a,b)\rightarrow \mathbb{R}^n is a parametrization of the curve C. It is then shown, that the integral is invariant under reparametrization. ... 0 Note that all your integrals are Gaussians in differences of positions at successive instants (x_k - x_{k-1}) so implement a change of integration variable from x_k \longrightarrow (x_k - x_{k-1}). You will have N-1 (straightforward) integrations to perform with x_0 and x_N held fixed. 5 1. Since x\gg p, we see that \sin(px) is highly oscillatory. In fact, the integral becomes$$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$modulo some factor of \pm2/i. Observe now this integral resembles \int f(p)\exp(g(p))\,\mathrm{d}p. We find the point ... 1 Looks like the book forgot a term in the integral$$\int_0^t \frac{F_0}{M} e^{-b t'}dt' = -\frac{F_0}{Mb} \left. \left( e^{-bt'}\right)\right|_0^t = \frac{F_0}{Mb}\left( 1 - e^{-bt} \right)$$For$t\rightarrow \infty$this goes to$F_0/bM$and for$t=0$it goes to$0\$.

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