Question

If 𝑭 = 𝟑𝒙𝒚𝒊 − 𝒚 𝟐 𝒋 evaluate ∫ 𝑭 ∙ 𝒅𝒓 where C is the curve in the xy plane, 𝒚 = 𝟐𝒙 𝟐 , from (0,0) to (1,2)

Answer 1

Given $\vec F=3xy\vec i-y^2\vec j$

$\vec r=x\vec i+y\vec j=>\vec {dr}=dx\vec i+dy\vec j$

$$F\cdot \vec dr=3xydx-y^2dy$$

$C$ is the part of the curve $y=2x^2$ from$(0, 0)$ to $(1, 2).$

$x(t)=t, y(t)=2t^2, t\in[0, 1]$

$dx=dt, dy=4tdt$

Then

$$\int_{C}\vec F\cdot \vec dr=\displaystyle\int_{0}^{1}(3t(2t^2)-(2t^2)^2(4t))dt$$

$$=\displaystyle\int_{0}^{1}(6t^3-16t^5)dt=\big[\dfrac{3t^4}{2}-\dfrac{8t^6}{3}\big]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=-\dfrac{7}{6}$$