Question

Find a function f such that ∇ f=(6yz,5xz,6xy−4z) f= Use f to eraluate ∫ C⟨5yz,5xz,5xy−4z⟩⋅dr= where C is the straight ine from (−1,−3,5) to (−1,−3,8) Venily that the Fundathental Theorem of Line integrals is conect by evaluating ∫ C(5yz,5zz;5zy−4z).

Answer 1

To find the scalar function f with a given gradient (6yz, 5xz, 6xy - 4z), integrate each component, yielding f(x, y, z) = 3xy(z·z - 1). The Fundamental Theorem of Line Integrals is used to evaluate the line integral along path C, resulting in the calculation of f at the endpoints (f at (8) minus f at (5)) when C is the path from (-1,-3,5) to (-1,-3,8).

Step-by-step explanation:

The question seeks to find a scalar function f such that its gradient ∇ f is equal to the vector field (6yz, 5xz, 6xy - 4z).

We know that the gradient of a function f in three dimensions is given by ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z). To find the function f, we can integrate each component of the gradient respectively:

1. ∂f/∂x = 6yz implies f = ∫ 6yz dx = 6xyz + g(y,z)
2. ∂f/∂y = 5xz suggests f = 5xyz + h(x,z)
3. ∂f/∂z = 6xy - 4z implies f = 3x·y·z² - 2z² + j(x,y)

Combining these partial integrals, we can deduce that f(x, y, z) = 3xy(z·z - 1), up to an arbitrary constant, can be a function whose gradient is (6yz, 5xz, 6xy - 4z).

Using this function f to evaluate the line integral over the path C, we apply the Fundamental Theorem of Line Integrals:

∫_C ⟨ (5yz, 5xz, 5xy - 4z) ⟩ · dr = f(– 1, – 3, 8) - f(– 1, – 3, 5) = 3(– 1)(– 3)(8· 8 - 1) - 3(– 1)(– 3)(5· 5 - 1)