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)`:


`(\partial f)/(\partial x)=4xy^3`

`\implies f=\displaystyle\int4xy^3\,\mathrm dx`

`f=2x^2y^3+g(y)`


`(\partial f)/(\partial y)=(\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`