Verify that the Divergence Theorem is true for the vector field F = 3x^2 i + 3xy
ID: 2878652 • Letter: V
Question
Verify that the Divergence Theorem is true for the vector field F = 3x^2 i + 3xyj + zk and the region E the solid bounded by the paraboloid z = 9 - x^2 - y^2 and the xy plane. To verify the Divergence Theorem we will compute the expression on each side. First compute tripleintegral_E div F dV div F = tripleintegral_E div F dV = integral^x_2_x_1 integral^y_2_y_1 integral^z_2_z_1 | dz dy dx| where x_1 = | y_1 = | z_1 = | x_2 = | y_2 = | z_2 = | tripleintegral_E div F dV = | Now compute double _S F middot dS| Consider S = P union D| where P| is the paraboloid and D is the disk. doubleintegral_P F middot dP = integral^x_2_x_1 integral^y_2_y_1 | dy dx| doubleintegral_D F middot dD = integral^x_2_x_1 integral^y_2_y_1 | dy dx| where x_1 = | y_1 = | x_2 = | y_2 = | double integral_P F middot dP = | double integral_D F middot dDExplanation / Answer
integral(E) div(F) dV = integral(E) (6x + 3x + 1) dV
x1 = -3 x2 = 3
y1 = -sqrt(9-x^2) y2 = sqrt(9-x^2)
z1 = 0 z2 = sqrt(9-x^2-y^2)
Using cylindrical coordinates,
= integral(r in [0,3], t in [0, 2pi], z in [0, 9-r^2]) (9r cos t + 1) (r dz dt dr)
= integral(r in [0,3], t in [0, 2pi]) (9r cos t + 1) r (9 - r^2) dt dr)
= integral(r in [0,3]) 2 pi * (9r - r^3) dr
= (2 pi) * (9/2r^2 - r^4/4) for r = 0 to 3
= (81/2) pi.
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Parameterize the surface via
r(u,v) = (u cos v, u sin v, 9 - u^2), u in [0,3], v in [0, 2pi]
r_u x r_v = (2u^2 cos v, 2u^2 sin v, u).
F = (3u^2 cos^2(v), 3u^2 cos v sin v, 9 - u^2)
So, integral(S) F * dS
= integral( u in [0,3], v in [0, 2pi]) (3u^2 cos^2(v), 3u^2 cos v sin v, 9 - u^2)*(2u^2 cos v, 2u^2 sin v, u) dv du
= integral(u in [0,3], v in [0, 2pi]) (6u^4 cos^3(v) + 6u^4 cos v sin^2(v)+ 9u - u^3) dv du
= integral(u in [0,3], v in [0, 2pi]) (6u^4 cos v + 9u - u^3) dv du
= integral(u in [0,3]) 2 pi * (9u - u^3) du
= (2 pi) * (9/2 u^2 - u^4/4) for u in [0, 3]
= 81/2 pi.
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