IIT JAM Follow
Vanshika Asked a Question
July 28, 2020 10:20 am 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
Reply(5)
Deepak singh
that's why we have put z = acosthta
• Vaishali sharma
dear see this
Reply(1)
Vaishali sharma
it's from geometry of sphere
• Ravi prakash yadav
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
Reply(0)
• Deepak singh
see attached
Reply(0)
• Ghost.
Because they convert Cartesian coordinate to spherical coordinate system and transformation is z=radius*cos(ø). wishes
Reply(0)
Head Office :
MPA 44, 2nd floor, Rangbari main Road Mahaveer Nagar II, Kota (Raj.) - 324005

Branch Office (Jaipur):
Shyam Tower, Plot No. F2, 6th Floor Amrapali Circle,
Vaishali Nagar, Jaipur, Rajasthan 302021

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