Let

F(x, y, z) = z tan−1(y^2)i + z^3 ln(x^2 + 10)j + zk.
Find the flux of F across S, the part of the paraboloid
x^2 + y^2 + z = 19
that lies above the plane
z = 3
and is oriented upward.

Thanks! STudying for a quiz...
*tan−1 is tan^(-1)

To find the flux of the vector field F across the given surface S, we can use the surface integral. The formula for the flux of a vector field across a surface is given by:

Flux = ∬S F · dS

where F is the vector field and dS is the differential area vector on the surface S.

In this case, we have F(x, y, z) = z tan^(-1)(y^2)i + z^3 ln(x^2 + 10)j + zk.

The surface S is defined by the part of the paraboloid x^2 + y^2 + z = 19 that lies above the plane z = 3 and is oriented upward.

To compute the flux, we need to determine the unit normal vector to the surface S. Since the surface S is defined by z = 19 - x^2 - y^2, the gradient of the surface gives us the normal vector:

n = ∇S = (∂z/∂x)i + (∂z/∂y)j + (-1)k

Let's find the partial derivatives:

∂z/∂x = -2x
∂z/∂y = -2y

Therefore, the unit normal vector n is:

n = (-2x)i + (-2y)j - k

Now, we can calculate the dot product F · dS:

F · dS = (z tan^(-1)(y^2)i + z^3 ln(x^2 + 10)j + zk) · ((-2x)i + (-2y)j - k)

Taking the dot product gives:

F · dS = -2xz tan^(-1)(y^2) - 2yz^3 ln(x^2 + 10) - z

To evaluate the surface integral, we need to parameterize the surface S. We can let:

x = r cosθ
y = r sinθ
z = 19 - r^2

where r ranges from 0 to √(19 - 3) = 4.

Next, we need to find the partial derivatives with respect to r and θ:

∂r/∂x = cosθ
∂r/∂y = sinθ
∂θ/∂x = -r sinθ
∂θ/∂y = r cosθ

The cross product between the partial derivatives will give us the differential area vector dS:

∂r/∂x × ∂θ/∂x = -r cosθ
∂r/∂y × ∂θ/∂y = -r sinθ

So, dS = (-r cosθ)i + (-r sinθ)j - rk

Finally, we can substitute the values of F, dS, and the parameterization into the flux formula:

Flux = ∬S F · dS
= ∫(θ=0 to 2π) ∫(r=0 to 4) (-2xrz tan^(-1)(y^2) - 2yrz^3 ln(x^2 + 10) - z)(-r cosθ)i + (-r sinθ)j - rk · (-r cosθ)i + (-r sinθ)j - rk dr dθ

Now, you can evaluate this double integral to find the flux of the vector field F across the given surface S.