In: Math
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 = 1
in the first octant, with orientation toward the origin
Solution :- we have F(x, y, z) = x i − z j + y k and S is the part of the sphere x2 + y2 + z2 = 1
Solving the equation of the sphere S for z we get z = √(1 - 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
where g(x, y) is the equation of S that is g(x, y) = √(1 - x^2 - y^2)
Now ∂g/∂x = -x/√(1 - x^2 - y^2) and ∂g/∂y = -y/√(1 - 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 ≤ 1, x ≤ 0, y ≤ 0.
Then ∫∫_S F • N dS = ∫∫_D F • [(∂g/∂x)i + (∂g/∂y)j - k] dA
= ∫∫_D (xi - zj + yk) • [{-x/√(1 - x^2 - y^2)}i + {-y/√(1 - x^2 - y^2)}j - k] dA
= ∫∫_D [-x^2/√(1 - x^2 - y^2) + yz/√(1 - x^2 - y^2) - y] dA
= ∫∫_D [-x^2/√(1 - x^2 - y^2) + y√(1 - x^2 - y^2)/√(64 - x^2 - y^2) - y] dA
= ∫∫_D [ -x^2/√(1 - x^2 - y^2) + y - y] dA
= ∫∫_D -[ x^2/√(1 - x^2 - y^2) dA
Now we change it to polar coordinates. The region D turns into 0 ≤ t ≤ π/2, 0 ≤ r ≤ 8 and using x = r cos(t) and y = r sin(t), we get
∫∫_D -[x^2/√(1- x^2 - y^2) ] dA
= (0 to π/2)∫ (0 to 1)∫ -[r cos(t)]^2/√(1 - r^2) ] r dr dt
= (0 to π/2)∫ (0 to 1)∫ [ -r^3*cos^2(t)/√(1 - r^2) ] dr dt
= -π/6