Question

`\int\limits^(6,4,8)_(1,0,-3) {x} \, dx + y\, dy - z\, dz`

Answer 1

Notice that
`(x,y,-z)` is a gradient field:

`\\abla f(x,y,z) = (x,y,-z) \implies \begin{cases} f_x = x \\ f_y = y \\ f_z = -z \end{cases}`

That is, there exists a scalar function
`f(x,y,z)` whose gradient is the given vector field. Solve for
`f`.

`\displaystyle \int f_x \, dx = \int x \, dx \implies f(x,y,z) = \frac12 x^2 + g(y,z)`

`f_y = g_y = y \implies \displaystyle \int g_y \, dy = \int y \, dy \implies g(y,z) = \frac12 y^2 + h(z)`

`f_z = h_z = -z \implies \displaystyle \int h_z \, dz = - \int z \, dz \implies h(z) = -\frac12 z^2 + C`

`\implies f(x,y,z) = \frac{x^2+y^2-z^2}2 + C`

By the gradient theorem, it follows that

`\displaystyle \int_((1,0,-3))^((6,4,8)) x \, dx + y \, dy - z \, dz = f(6,4,8) - f(1,0,-3) = \boxed{-2}`