Final answer:
To find the work done by the vector field F=3xyi+8y²j along the path y=2x², we parametrize the path using x, substitute into the x-component of the vector field, and then evaluate the integral of that component over the interval [0,1], leading to a result of 1.5 joules.
Step-by-step explanation:
The student is asking about how to evaluate the line integral of a vector field F along a path C. To do this, we must parametrize the path and then compute the integral of the dot product of the vector field and the differential displacement vector dx over the given path.
In this case, the path C is defined by y=2x², which joins the points (0,0) and (1,2). The vector field is given by F=3xyi+8y²j. To parametrize the path using a single variable, we can use x as our parameter, then y will be 2x². Thus, the path C can be described parametrically as (x, 2x²) where x ranges from 0 to 1.
The dot product F ⋅ dx only includes the x-component of the vector field since dx in this context represents an infinitesimal displacement in the x-direction. So we consider the x-component of the vector field, which is 3xy, and substitute y = 2x² into it, yielding 6x³. Consequently, the integral becomes ∫ 6x³ dx over the interval [0,1]. Calculating this integral yields the work done along the path C:
Substitute y = 2x² into the x-component of F to get the integrand 6x³.
Evaluate the integral ∫ 6x³ dx from 0 to 1.
Calculate the definite integral, which gives 6/4, or 1.5 J (joules) as the work done by the force vector field F along the path C from (0,0) to (1,2).