Question

Hello, can please help me with this problem? Can you please do the problem without using parametric surfaces! Please do the problem using the definition of surface integrals over vector fields!

F. dS where F = i + zj+ 6e k and S is the portion of the sphere of radius 3 with 0, y 0 and z 0 oriented inward (i.e. towards 4. Evaluate S the origin). [Solution]

Answer 1

F. dS where F = 7+ zj +6z k and S is the portion 'radius 3 with a < 0, y 2 0 and z 0 oriented inward the sphere 4. Evaluate (i.e. towards S the origin). [Solution] 3 2 1 4 X The surface is sphere we will need to use the parametric -1 -3 representation of the surface. This is, F (0, p)=3 cos pi+3 sinp cos 0j+3sinp sin0k Since we are working on the hemisphere here are the limits on the parameters that we'll need to use. 0 Next, we need to determine rg X Ta Here are the two individual vectors and the cross product fo (0, p) = -3sin p sin 0 j +3 sin p cos 0 k F (0, )=-3 sin p i +3 cos pcos 0j+3 cos p sin0 k Therefore, 0 -3 sinp sin 0 3 sin p cos 3 cos p cos 3 cos p sine 9 sin p cos pcos20i-9sino cos psin?0i -9sin'p cos 0 j -9sin'ysin0 k -3 sinp 9 sin pcos p (sin20 + cos20) i 9sin2p cos 0 j -9sin2p sin 0 k 9sin2 cos0j =-9 sinp cos p i 9sin2sin0k The vector will have a positive component and as we noted above in order for this to point away from the enclosed area we will need the component to be negative Therefore, we will need to use the following vector for the unit normal vector. 9 sin p cos p i 9sin2p cos 0j +9sin2p sin0 k Fox f|| Next we'll need F (7(0, ))= i+9sin2psin0j+36 sinp cos p k Now we can do the integral. lJ P Foxl dA F.dS = D S 27 (9 sin p cos p+81 sin sin0 cos 0+ 324 sin3p sin 0 cos p) dp de 24S %> & + 224 sin"& Sind 2- sin"(x)cbc = 37 2 9,243% sino s E ニー Steps - 9 6 sin (ax) cac= :) cos (x) + 2 Compute the boundaries: -0-81 (1-1) = t18