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
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  • Deepak singh Best Answer
    see attached
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    Deepak singh
    that's why we have put z = acosthta
  • Vaishali sharma thankyou
    dear see this
    • cropped7168143134135399832.jpg
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    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
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  • Deepak singh
    see attached
    • cropped43306764.jpg
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  • Ghost.
    Because they convert Cartesian coordinate to spherical coordinate system and transformation is z=radius*cos(ø). wishes
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