sproblems9 - Dr H Khanal Spring 2008 MA 441 – Solved...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Dr. H. Khanal Spring 2008 MA 441 – Solved Problems #10. Stokes’s Theorem 1. Integrate the surface integral directly. Then check the result by integrating the corresponding line integral in the Stokes’s theorem. Z Z S ∇ × F · n dA = I C F · r ( s ) ds Sol. Here F = [ y 2 ,- x 2 , 0] and the surface S is given by x 2 + y 2 ≤ 4 , y ≥ , z = 0. We can express the surface parametrically as r ( u,v ) = [ u cos v, u sin v, 0] , ≤ u ≤ 2 , ≤ v ≤ π. We calculate the normal vector using the formula N = r u × r v . r u = [cos v, sin v, 0] r v = [- u sin v, u cos v, 0] N = r u × r v = fl fl fl fl fl fl fl fl i j k cos v sin v- u sin v u cos v fl fl fl fl fl fl fl fl = [0 , , u ] . ∇ × F = fl fl fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z y 2- x 2 fl fl fl fl fl fl fl fl = [0 , ,- 2 x- 2 y ] . = ⇒ ∇ × F ( u,v ) = [0 , ,- 2 u (cos v + sin v )] , ∇ × F · N = [0 , ,- 2 u (cos v + sin v )] · [0 , , u ] =- 2 u 2 (cos v + sin v ) and the surface integral is Z Z...
View Full Document

This note was uploaded on 06/03/2008 for the course MA 441 taught by Professor Kaba during the Fall '08 term at Embry-Riddle FL/AZ.

Page1 / 2

sproblems9 - Dr H Khanal Spring 2008 MA 441 – Solved...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online