Suppose \nabla f (x, y) = 3 y \sin(xy) \vec{i} + 3 x \sin(xy)\vec{j}, \vec{f} = \nabla f(x, y), and c is the segment of the parabola y = 3 x^2 from the point (1,3) to (4,48). then

I'll assume you're supposed to compute the line integral of over the given path . By the fundamental theorem of calculus,

so evaluating the integral is as simple as evaluting at the endpoints of . But first we need to determine given its gradient.

We have

Differentiating with respect to gives

and we end up with

for some constant . Then the value of the line integral is .

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)