![\large \\ \text{ Flux of F across surface S is given by } \int \int _S F.\hat n dS\\\\ \text{Divergence theorem : It associates flux of vector field through a closed surface to divergence over region bounded by surface}\\ \text{Let D is solid region bounded by surface S then }\\ \int \int _S F.\hat n dS=\int\int\int_D div(F).dV\\\\ div(F)=0+1+1=2\\\\ \begin{align*} Flux&=\int_0^{2\pi}\int_0^{\pi}\int_0^{\phi}div(F)\rho^2\sin \phi d\rho d\phi d\theta\\\\ &=\frac{2}{3}\int_0^{2\pi}\int_0^{\pi}\phi^3\sin \phi d\phi d\theta\\\\ &=\frac{2}{3}\int_0^{2\pi}\left[-\phi^3 \cos\phi +3\phi^2\sin\phi+6\phi\cos\phi-6\sin\phi \right ]_0^{2\pi}\\\\ &=\frac{2(\pi^3-6\pi)}{3}2\pi\\\\ &=\frac{4(\pi^4-6\pi^2)}{3} \end{align*}](http://img.homeworklib.com/questions/04846d90-f2f2-11ea-8960-379f661da402.png?x-oss-process=image/resize,w_560)
9. X-rays have intensity and direction that are given by a vector field F(x, y, z)...
(1) Let F denote the inverse square vector field (axr, y, z) F= (Note that ||F 1/r2.) The domain of F is R3\{(0, 0, 0)} where r = the chain rule (a) Verify that Hint: first show that then use (b) Show that div(F 0. (c) Suppose that S is a closed surface in R3 that does not enclose the origin. Show that the flux of F through S is zero. Hint: since the interior of S does not contain...
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rough the surface 4. o pm) What is the value of the flux of the vector field F(x,y)j+z ioriented with upward- pointing normal vector? (A) 0 (B) 2n/3 (C) π (D) 4T/3 (E) 2π Use Stokes, Theorem to evaluateⅡcurl F.dS, where F(x, y, z)-(x2 sin Theorem to evaluate Jceun F'.asS , where Fl.e)(', ») and 5. (5pts.) F,y, sin z, y', xy) and s is the part of the paraboloid : -...
Il Evaluate the surface integral F.ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = y) - zk, S consists of the paraboloid y = x2 + 22,0 Sys1, and the disk x2 + z2 s 1, y = 1. Evaluate the surface integral F.ds for the given vector field F and the oriented surface S....
Evaluate the surface integral F dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) -xi yj+3 k S is the boundary of the region enclosed by the cylinder x2 + z2-1 and the planes y 0 and x y 2
Evaluate the surface integral F dS for the given vector field F and the oriented surface...
Evaluate the surface integral | Fids for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the JS positive (outward) orientation. F(x, y, z) = y i + (z - y)j + xk S is the surface of the tetrahedron with vertices (0, 0, 0), (4, 0, 0), (0, 4, 0), and (0, 0, 4)
xi+ yj + zk 3. Given the vector field in space F(x, y, z) = or more conveniently, (.x2 + y2 + 22)3/2 1 Fr) where r = xi + yj + zk and r= ||1|| = x2 + y2 + x2 (instead of p) 73 r (a) [10 pts) Find the divergence of F, that is, V.F. (b) (10 pts] Directly evaluate the surface integral [/F F.Nds where S is the unit sphere 22 + y2 + z2 1...
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xzey i − xzey j + z k S is the part of the plane x + y + z = 7 in the first octant and has upward orientation.
Evaluate the surface integral F.ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation F(x, y, z) = yi - xj + Szk, S is the hemisphere x2 + y2 + z2 = 4, z 20, oriented downward -8751 x
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = x i − z j + y k S is the part of the sphere x2 + y2 + z2 = 36 in the first octant, with orientation toward the origin
Evaluate the surface integral ∫∫S F·ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xyl + yzj + zxk S is the part of the paraboloid z = 2 = x2 - y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation_______