Question

83

Answer 1

1. Observe that

$\nabla \dfrac{x^2y^2w^2}2 = \langle xy^2w^2, x^2yw^2, x^2y^2w\rangle$

is a gradient field, so the gradient theorem holds and the integral in question is indeed path-independent. Its value is

$\dfrac{x^2y^2w^2}2\bigg|_{x=1,y=3,w=2}^{x=2,y=4,w=1} = 32 - 18 = \boxed{14}$

2. $dz$ is an exact differential if we can find a scalar function $z=f(x,y)$ such that

$\dfrac{\partial f}{\partial x} = 3x^2 (x^2+y^2) = 3x^4 + 3x^2y^2$

$\dfrac{\partial f}{\partial y} = 2y(x^3+y) = 2x^3y + 2y^2$

Integrating both sides of the first equation with respect to $x$ yields

$f(x,y) = \dfrac35 x^5 + x^3 y^2 + g(y)$

Differentiating with respect to $y$ gives

$\dfrac{\partial f}{\partial y} = 2x^3y + \dfrac{dg}{dy} = 2x^3y + 2y^2 \\\\ \implies \dfrac{dg}{dy} = 2y^2 \implies g(y) = \dfrac23y^3 + C$

and we ultimately find

$f(x,y) = \boxed{z = \dfrac35 x^5 + x^3y^2 + \dfrac23 y^3 + C}$

(We can also use the same method here to determine the scalar function in part (1).)

Then the integral is path-independent, and its value is

$f(2,1) - f(1,2) = \dfrac{418}{15} - \dfrac{149}{15} = \boxed{\dfrac{269}{15}}$