- #1
cronxeh
Gold Member
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Hey can you guys check my answer.
Question: Use the Divergence Theorem to calculate the Flux of the vector field F(x,y,z)=xi + y^2j - zk through the unit sphere centered at the origin with the outward orientation
Solution: div(F) = 1 + 2y - 1 = 2y
Flux = [tex]\int_{W} div(F) dV = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 2y \ dxdydz = \int_{0}^{1} \int_{0}^{1} 2y \ dydz = \int_{0}^{1} 1 \ dz = 1 [/tex]
So for another method would I have to use [tex]r=x^2 + y^2 + z^2[/tex] obtain the [tex]dr=2x + 2y + 2z[/tex] and do [tex] \int_{R} F dr = \int_{S} (xi + y^2j -zk)(2x+2y+2z)dA=\int_{S} (2x^2 + 2y^3 - 2z^2)dA = \int_{0}^{1} \int_{0}^{1} 2/3 + 2y^3 - 2z^2 \ dydz = \int_{0}^{1} 2/3 + 1/2 - 2z^2 dz = 2/3 + 1/2 - 2/3 = \frac{1}{2} [/tex]
Is this method incorrect or was the first one incorrect? Are they both wrong?
Question: Use the Divergence Theorem to calculate the Flux of the vector field F(x,y,z)=xi + y^2j - zk through the unit sphere centered at the origin with the outward orientation
Solution: div(F) = 1 + 2y - 1 = 2y
Flux = [tex]\int_{W} div(F) dV = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 2y \ dxdydz = \int_{0}^{1} \int_{0}^{1} 2y \ dydz = \int_{0}^{1} 1 \ dz = 1 [/tex]
So for another method would I have to use [tex]r=x^2 + y^2 + z^2[/tex] obtain the [tex]dr=2x + 2y + 2z[/tex] and do [tex] \int_{R} F dr = \int_{S} (xi + y^2j -zk)(2x+2y+2z)dA=\int_{S} (2x^2 + 2y^3 - 2z^2)dA = \int_{0}^{1} \int_{0}^{1} 2/3 + 2y^3 - 2z^2 \ dydz = \int_{0}^{1} 2/3 + 1/2 - 2z^2 dz = 2/3 + 1/2 - 2/3 = \frac{1}{2} [/tex]
Is this method incorrect or was the first one incorrect? Are they both wrong?