Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S.?
In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
1)F(x, y, z) = xy i + yz j + zx k
S is the part of the paraboloid z = 6 − x^2 − y^2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.
2)F(x, y, z) = xze^y i − xze^y j + z k
S is the part of the plane x + y + z = 5 in the first octant and has downward orientation.
3)F(x, y, z) = xz i + x j + y k
S is the hemisphere x^2 + y^2 + z^2 = 9, y ≥ 0, oriented in the direction of the positive y-axis.
I can always get to the point where F(r(x,y,)) but I'm not sure where do go on from there.
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1) Using rectangular coordinates,
∫∫s F · dS
= ∫∫ <xy, yz, zx> · <-z_x, -z_y, 1> dA
= ∫∫ <xy, y(6 - x^2 - y^2), x(6 - x^2 - y^2)> · <2x, 2y, 1> dA
= ∫(x = 0 to 1) ∫(y = 0 to 1) [2x^2 y + 2y^2 (6 - x^2 - y^2) + x(6 - x^2 - y^2)] dy dx
This is straightforward to integrate from here.
(answer 1133/180)
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2) Write z = 5 - x - y, and note that the region of integration is between x + y = 5 and the x,y axes. Then, proceeding as #1:
∫∫s F · dS
= ∫∫ <xze^y, -xze^y, z> · <-z_x, -z_y, 1> dA
= ∫∫ <x(5 - x - y)e^y, -x(5 - x - y)e^y, (5 - x - y)> · <1, 1, 1> dA
= ∫(x = 0 to 5) ∫(y = 0 to 5 - x) (5 - x - y) dy dx
= ∫(x = 0 to 5) [(5 - x)y - y^2/2] {for y = 0 to 5 - x} dx
= ∫(x = 0 to 5) (1/2) (5 - x)^2 dx
= (-1/6)(5 - x)^3 {for x = 0 to 5}
= 125/6.
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3) This is similar to 1, but we need a slightly different rectangular parameterization.
Write r(x, z) = <x, y(x, z), z>
==> r_x x r_y = <y_x, -1, y_z>.
Since we want the normal to point in the positive y-direction, we negate our result:
n = <-y_x, 1, y_z>.
Since y = √(9 - x^2 - z^2), we have
∫∫s F · dS
= ∫∫ <xz, z, y> · <-y_x, 1, y_z> dA
= ∫∫ <xz, z, √(9 - x^2 - z^2)> · <x/√(9 - x^2 - z^2), 1, z/√(9 - x^2 - z^2)> dA
= ∫∫ [x^2 z/√(9 - x^2 - z^2) + 2z] dA
Noting that the region of integration is x^2 + z^2 ≤ 9, converting to polar coordinates with
x = r cos θ and z = r sin θ yields
∫(r = 0 to 3) ∫(θ = 0 to 2π) [(r^3 cos^2(θ) sin θ/√(9 - r^2) + 2r sin θ] * r dθ dr
= 0.
I hope this helps!