Find the flux of the given vector field...

Find the flux of the given vector field F across the upper hemisphere x^2 + y^2 + z^2 = a^2, z >= 0. Orient the hemisphere with an upward-pointing normal.
a. F = yj
b. F = -yi + xj - k

Please help

Writing the equation of the upper hemisphere as z = √(a^2 - x^2 - y^2),
∫∫s F · dS
= ∫∫ F · <-z_x, -z_y, 1> dA, via cartesian coordinates
= ∫∫ F · dA

Moreover, the region of integration is the interior of x^2 + y^2 = a^2.
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a) Letting F = <0, y, 0>, we obtain
∫∫ [y^2/√(a^2 - x^2 - y^2)] dA
= ∫(θ = 0 to 2π) ∫(r = 0 to a) [r^2 sin^2(θ) / √(a^2 - r^2)] * r dr dθ, via polar coordinates
= ∫(θ = 0 to 2π) sin^2(θ) dθ * ∫(r = 0 to a) r^3 dr / √(a^2 - r^2)
= ∫(θ = 0 to 2π) (1/2)(1 - cos(2θ)) dθ * ∫(r = 0 to a) r^3 dr / √(a^2 - r^2)
= π * ∫(r = 0 to a) r^3 dr / √(a^2 - r^2)
= π * ∫(t = 0 to π/2) (a^3 sin^3(t)) * (a cos t dt) / (a cos t), letting r = a sin t
= πa^3 * ∫(t = 0 to π/2) sin^3(t) dt
= πa^3 * ∫(t = 0 to π/2) (1 - cos^2(t)) sin t dt
= πa^3 * (-cos t + cos^3(t)/3) {for t = 0 to π/2}
= 2πa^3/3.
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b) Letting F = <-y, x, -1>, we obtain
∫∫ -1 dA
= -(Area inside x^2 + y^2 = a^2)
= -πa^2.

I hope this helps!

Using spherical coordinates, you will see that:

x = acos(φ)sin(θ)
y = asin(φ)sin(θ)
z = acos(θ)

Now performing the cross product between the tangential vectors Tφ and Tθ, you will see that you get
(-a²sin²θ*cosφ) i - (a²sin²θ*sinφ)j - (a²sinθcosθ)k

Now dotting this with F = yj = (asinφsinθ)j and integrating:

∫[0, π] ∫[0, π/2] -a³sin³(θ)sin²φ dθ dφ

= -2a³/3 * ∫[0, π] sin²φ dφ

= -2a³/3*[π/2] = -πa³/3