Soumi Mukherjee Asked a Question
January 4, 2020 8:33 pmpts 15 pts
Calculate the surface integral of V=2xz x + (X +2) y + y(Z -3)z over five sides (excli bottom) of the cubical box shown in figure. Taking the sides one at a time: () (i) Ex. Sol. x = 2, d = dy dz , v.dš = 2xzdy dz = 4z dy dz So, Jv.ds = 0 x = 0, ds = -dy dz å, v.dš= -2xzdy dz =0, so Jv.ds = 0 (v) ZA (i) (ii) (iv) X 2, dš = dx dz ý, vdš=(x+2) dx dz so V.ds =|(x +2) dx dz = 12 (i) y y = 0, ds= -dx dz ý, vdš= -(x+2) dx dz so vds-jx+2) dx| dz =-12 () z= 2, dš = dxdy ż, vdš = ydxdy, so v.ds = | dxydy=4 (v) Evidently the total flux is v.ds =16+0+12-12+4 =20 surface
  • 1 Answer(s)
  • 0 Likes
  • 1 Comments
  • Shares
  • Dhairya sharma thankyou
    it may be a printing mistake because it's yz plane and value of y and 2 are changing from 0 to 2 . so u r right it's 16. and at the end of the solution it's also clear that value o...
    Show more
    Likes(0) Reply(1)
    Soumi Mukherjee
    yes..eventually I saw the '16' in the final step
Head Office :
MPA 44, 2nd floor, Rangbari Main Road,
Mahaveer Nagar II, Kota (Raj.) – 324005

Corporate Office:
212, F-1, 2nd Floor, Evershine Tower,
Amrapali Marg,
Vaishali Nagar, Jaipur (Raj.) – 302021

Mail: info@eduncle.com
All Rights Reserved © Eduncle.com