profile-img
Vanshika Asked a Question
July 28, 2020 10:20 ampts 30 pts
0 =k(1-4) -3k x X ay 0z (B) Here Vx V= 4y 0 2z Since the surface is a sphere with centre at the origin, f is its normal, then n1a a ix ix +jy +kz (V xv) h3k3 a a We want to evaluated-3(z/a) do over the hemisphere, In spherical coordirs 2n n/2 aCos a sin 0 de do-3a d a cos.o a sin 0 cos 0 do -0 0-0 A =- 3a2.2t. -3rma
  • 3 Answer(s)
  • 0 Likes
  • 5 Comments
  • Shares
  • Deepak singh Best Answer
    see attached
    • cropped896655595.jpg
    Likes(1) Reply(5)
    Deepak singh
    that's why we have put z = acosthta
  • Vaishali sharma thankyou
    dear see this
    • cropped7168143134135399832.jpg
    Likes(1) Reply(1)
    Vaishali sharma
    it's from geometry of sphere
  • Ravi prakash yadav thankyou
    in spherical coordinate z=acos(theta) and da=a^2sin(theta)d(theta)d(phi). we simply put z=acos(theta) and da=a^2sin(theta)d(theta)d(phi) in original equation
    Likes(0) Reply(0)
  • Deepak singh
    see attached
    • cropped43306764.jpg
    Likes(0) Reply(0)
  • Ghost.
    Because they convert Cartesian coordinate to spherical coordinate system and transformation is z=radius*cos(ø). wishes
    Likes(0) Reply(0)