[Émilie.du]

Compute the line integral \( \displaystyle \int_{C}(x^2-2xy)dx+(y^2-2xy)dy \) along the curve \( C=\left\{{(x,y)\in \mathbb{R^2}: x\in[-1,1], y=x^2}\right\} \).

Hi, I need help. Please.
How do I do this using differential forms?

Regards.

[Luis Fuentes]

Hi

Compute the line integral \( \displaystyle \int_{C}(x^2-2xy)dx+(y^2-2xy)dy \) along the curve \( C=\left\{{(x,y)\in \mathbb{R^2}: x\in[-1,1], y=x^2}\right\} \).

Hi, I need help. Please.
How do I do this using differential forms?

Regards.

Parameterize the curve \( C \):

\( (x(t),y(t))=(t,t^2) \) with \( t\in [-1,1] \)

then \( dx=dt,\quad dy=2tdt. \)

The integral becomes:

\( \displaystyle\int_{-1}^1((t^2-2t^3)+(t^4-2t^3)(2t))dt \)

Best regards.

[Émilie.du]

Thanks.
I think this result is similar:
Let \( \omega=(x^2-2xy)dx+(y^2-2xy)dy \Rightarrow C^*(\omega)=C^*((x^2-2xy)dx)+C^*((y^2-2xy)dy)=((x^2-2xy)\circ C)d(x\circ C)+((y^2-2xy)\circ C)d(y\circ C) \)
                                                                             \( =(x^2-2x^3)d(x)+(x^4-2x^3)d(x^2)=(x^2-2x^3)dx+(x^4-2x^3)(2xdx) \)
                                                                             \( =(2x^5-4x^4-2x^3+x^2)dx \)
\( \Rightarrow \displaystyle\int_{C}\omega=\displaystyle\int_{-1}^{1}(2x^5-4x^4-2x^3+x^2)dx=\dfrac{x^6}{3}-\dfrac{4x^5}{5}-\dfrac{x^4}{2}+\dfrac{x^3}{3} |_{x=-1}^{x=1}=\dfrac{-14}{15} \)

[Luis Fuentes]

Hola

Thanks.
I think this result is similar:
Let \( \omega=(x^2-2xy)dx+(y^2-2xy)dy \Rightarrow C^*(\omega)=C^*((x^2-2xy)dx)+C^*((y^2-2xy)dy)=((x^2-2xy)\circ C)d(x\circ C)+((y^2-2xy)\circ C)d(y\circ C) \)
                                                                             \( =(x^2-2x^3)d(x)+(x^4-2x^3)d(x^2)=(x^2-2x^3)dx+(x^4-2x^3)(2xdx) \)
                                                                             \( =(2x^5-4x^4-2x^3+x^2)dx \)
\( \Rightarrow \displaystyle\int_{C}\omega=\displaystyle\int_{-1}^{1}(2x^5-4x^4-2x^3+x^2)dx=\dfrac{x^6}{3}-\dfrac{4x^5}{5}-\dfrac{x^4}{2}+\dfrac{x^3}{3} |_{x=-1}^{x=1}=\dfrac{-14}{15} \)

Ok. It is the same solution replacing \( t \) by \( x \).

Best regards.