Question

Find ∫c f*dr where f(x,y) = 〈 4xy3 , 6x2y2 〉 and c is a sequence of straight lines from (0,0) to (1,3) to (4,7) to (9,5) to (2,1

Answer 1

First note that $\mathbf f(x,y)=\langle4xy^3,6x^2y^2\rangle$ is continuous, which means that there is some scalar function $f(x,y)$ whose gradient is $\mathbf f(x,y)$. In other words, $\mathbf f(x,y)$ is conservative, and so the gradient theorem applies here, which means the value of the line integral is given by $f(2,1)-f(0,0)$.

So we look for such a function $f(x,y)$:

$\dfrac{\partial f}{\partial x}=4xy^3$
$\implies f=\displaystyle\int4xy^3\,\mathrm dx$
$f=2x^2y^3+g(y)$

$\dfrac{\partial f}{\partial y}=\dfrac{\partial(2x^2y^3+g(y))}{\partial y}$
$6x^2y^2=6x^2y^2+g'(y)$
$0=g'(y)$
$\implies g(y)=C$

$\implies f(x,y)=2x^2y^3+C$

Thus by the gradient theorem, the value of the line integral is

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=f(2,1)-f(0,0)=8$