Question

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.

F(x, y, z) = x i − z j + y k

S is the part of the sphere

x² + y² + z² = 9

in the first octant, with orientation toward the origin

Answer 1

Solution :- we have x^2+y^2+z^2 =9

Solving the equation of the sphere S for z we get z = √(9 - x^2 - y^2)

Since we're considering the first octant only, all the points are above the origin, thus S is oriented downward. Then the flux integral is:

∫∫_S F • N dS = ∫∫_D F • [(∂g/∂x)i + (∂g/∂y)j - k] dA

Here, we have F = xi - zj + yk and g(x, y) = z = √(9 - x^2 - y^2) ==> ∂g/∂x = -x/√(9 - x^2 - y^2) and ∂g/∂y = -y/√9 - x^2 - y^2). D is the region of the projection of S onto the xy-plane, which in this case is x^2 + y^2 ≤ 9, x ≤ 0, y ≤ 0.

So we have ∫∫_S F • N dS = ∫∫_D F • [(∂g/∂x)i + (∂g/∂y)j - k] dA

= ∫∫_D (xi - zj + yk) • [{-x/√(9 - x^2 - y^2)}i + {-y/√(9 - x^2 - y^2)}j - k] dA

= ∫∫_D [-x^2/√(9- x^2 - y^2) + yz/√(9 - x^2 - y^2) - y ]dA

= ∫∫_D [-x^2/√(9 - x^2 - y^2) + y√(9 - x^2 - y^2)/√(9 - x^2 - y^2) - y] dA

= ∫∫_D [-x^2/√(9 - x^2 - y^2) + y - y ] dA

= ∫∫_D [-x^2/√(9 - x^2 - y^2) ] dA

Now we switch to polar coordinates. The region D turns into 0 ≤ t ≤ π/2, 0 ≤ r ≤ 3 and using x = r cos(t) and y = r sin(t), we get

∫∫_D -x^2/√(9 - x^2 - y^2) dA

= (0 to π/2)∫ (0 to 3)∫ -[r cos(t)]^2/√(9 - r^2) ]r dr dt

= (0 to π/2)∫ (0 to 3)∫ -r^3*cos^2(t)/√(9 - r^2) dr dt

= -9 pi/2