To compute the line integral of ∇f along the curve c, we need to find the dot product of the gradient vector of f and the tangent vector of the curve c, and then integrate it over the interval [0, 1].
First, let's find the gradient vector of f:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)
∂f/∂x = y^3z^2
∂f/∂y = 3xy^2z^2
∂f/∂z = 2xy^3z
Therefore, ∇f = (y^3z^2, 3xy^2z^2, 2xy^3z).
Next, let's find the tangent vector of the curve c:
r(t) = (et cos(t^2 + 1), ln(t^2 + 1), 1/(t^2 + 1))
To find the tangent vector, we take the derivative of r(t) with respect to t:
r'(t) = (et cos(t^2 + 1) - 2et^3 sin(t^2 + 1), 2t/(t^2 + 1), -2t/(t^2 + 1)^2)
Now, we calculate the dot product of ∇f and r':
∇f · r' = y^3z^2(et cos(t^2 + 1) - 2et^3 sin(t^2 + 1)) + 3xy^2z^2(2t/(t^2 + 1)) + 2xy^3z(-2t/(t^2 + 1)^2)
To evaluate this dot product along the curve c, we substitute the corresponding values of x, y, and z from the curve into the expression.
x = et cos(t^2 + 1)
y = ln(t^2 + 1)
z = 1/(t^2 + 1)
Substituting these values, we get:
∇f · r' = ln(t^2 + 1)^3(1/(t^2 + 1))^2(et cos(t^2 + 1) - 2et^3 sin(t^2 + 1))
+ 3et cos(t^2 + 1)ln(t^2 + 1)^2(1/(t^2 + 1))^2
- 2et^3 sin(t^2 + 1)ln(t^2 + 1)^3(1/(t^2 + 1))
Now, we integrate this expression over the interval [0, 1]:
Line integral of ∇f along c = ∫[0,1] ∇f · r' dt
Evaluate this integral using the given limits.