Question

the answer is

why domain of p is 【0，1】in here

And the answer continuous like this , the second question is why the surface integral over the top z=2?

And how to get the surface integral over the top of the cylinder and the surface integral over the bottom of the cylinder?

over the cylindery20 2. 1) First we compute the volume integral using cylinder coordinates r-ρ cos φ, y ρ sino, 2 = ρ with volume element dV = ρdpd2db. We need Then we compute (6-42+ 2pa cos φ)pdpdzdo z cOS cOS Next we evaluate the surface integral over the curved surface of the cylinder parame- terized as The tangent vectors are T,- sin n't cos and T, OF-k and the outward pointing normal vector then results to T, x T: cosor+sino, Using the above parameterization, we express F as a function of θ and and compute --Sin J0 J0 J0 The surface integral over the top of the cylinder is parameterized as with tangent vectors T,-器=-2sin φ→+ρ cos01 and T,-器= cos n't sin φ]. The outward, the is upward, pointing normal vector is then Tf T,-ρk Using the above parameterization, we express F as a function of θ and r and compute Therefore 1 2 49 The surface integral over the bottom of the cylinder is parameterized as with tangent vectors Τφ 5-psin n't ρ coso,and Τρ ρ = dos φ→+ sino, The outward, that is downward, pointing normal vector is then T, xTp--pk. Using the above parameterization, we express F as a function of θ and r and compute 一-p", S111 Therefore 1 2π Adding up all the contributions from the surface integrals gives precisely 4T, that is the value of the volume integral

Answer 1

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