Show that if the vector field f = pi + qj + rk is conservative and p, q, r have continuous first-order partial derivatives, then the following is true. ∂p ∂y = ∂q ∂x ∂p ∂z = ∂r ∂x ∂q ∂z = ∂r ∂y . since f is conservative, there exists a function f such that f = ∇f, that is, p, q, and r are defined as follows. (enter your answers in the form fx, fy, fz.)

The field F has a continuous partial derivative on R.

Step-by-step explanation:

For the field F has a continuous partial derivative on R, fxy must be equal to fyx and since our field F is ∇f,

∇f = fxi + fyj + fzk.

Comparing the field F to ∇f since they at equal, P = fx, Q = fy and R = fz

Since P is fx therefore;

∂P ∂y = ∂ ∂y( ∂f ∂x) = ∂2f ∂y∂x

Similarly,

Since Q is fy therefore;

∂Q ∂x = ∂ ∂x( ∂f ∂y) = ∂2f ∂x∂y

Which shows that ∂P ∂y = ∂Q ∂x

The same is also true for the remaining conditions given

the answer is b i just did that

step-by-step explanation:

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !