Krdytkyu

up vote 0 down vote favorite Let $r > 0$ . Suppose $fin C^2 (overline{Bbb R^d setminus B(0,r)})$ , where $B(0,r) := { xin Bbb R^d : Vert x Vert leq r }$ . This means $fin C^2 (Bbb R^d setminus B(0,r))$ and has a continuous extension to the boundary of $B(0,r)$ . Take $R > r$ . Now define $$varphi (x) := begin{cases} exp left( frac{R^2 - Vert x Vert^2}{r^2 - Vert x Vert^2} right) &: r < Vert x Vert < R\ 1 &: Vert x Vert geq R \ 0 &: Vert x Vert leq r end{cases}$$ Does it hold that $varphi in C^infty(Bbb R^d)$ and $varphi f in C^2(Bbb R^d)$ ? And if so, how do I show it? derivatives smooth-functions share | cite | improve this question asked Nov 16 at 1

up vote 0 down vote favorite 1 I am asked to evaluate the following integral: $$intint text{curl} vec{F} cdot dvec{S}$$ where $F = (y,-x,zx^3y^2)$ and $S$ is the surface given by $x^2 + y^2 + 3z^2 = 1$ with $z leq 0$ . I did not have any problem with any other exercises of this kind. But this one is hard. Any hint would be appreciated. surface-integrals stokes-theorem share | cite | improve this question asked Nov 16 at 19:10 TheNicouU 171 1 11