Question

Evaluate the differential form of the vector field.\( \int_Cx^2ydx-xydy \) ; C is the curve with equation \( y^2=x^3 \) ,from (1,-1) to (1,1)

 

Answer 1

Solution \( \)

we have \( \int_Cx^2ydx-xydy \) ; C is the curve with equation 

\( y^2=x^3 \) ,from (1,-1) to (1,1)

1st Method:

We have  \( y^2=x^3=>x=y^{3/2}=>dx=\frac{2}{3y^{1/3}}dy \)

\( =>\int_Cx^2ydx-xydy=\int_C\bigg((y^{4/3})\frac{2y}{3y^{1/3}}dy -{y}^{2/3}ydy\bigg) \)

                                     \( =\int_C\bigg(\frac{2}{3}y^2-y^{5/3} \bigg)dy \) where \( -1\leq y \leq 1 \)

                                    \( =\int_{-1}^1\bigg(\frac{2}{3}y^2-y^{5/3} \bigg)dy=4/9 \)

2nd Methot:

Let x=t =>dx=dt

And then =>\( y=t^{3/2}=>dy=\frac{3}{2}t^{1/2} \) where \( t\in [-1,1] \)

\( =>\int_Cx^2ydx-xydy=\int_{-1}^1\bigg(t^{7/2}-\frac{3}{2}t^3\bigg)dt=4/9 \)

  

Answer

Therefore, \( \int_Cx^2ydx-xydy =4/9 \)\( \)