Solve the following question about line integral

Question

Given, $$\vec{F} = \nabla (4x^2+3y^4)$$

Find $$\int_{C} \vec{F}.d\vec{r}$$ where $C$ is the quarter of circle $x^2+y^2=1$ in first quadrant, rotated counterclockwise.

Answer 1

Use the general result $$ \int_C \nabla V\cdot d\mathbf r= V(\mathbf x_1)-V(\mathbf x_0) $$ where $\mathbf x_0$ is the initial point of $C$ and $\mathbf x_1$ the final point.

Answer 2

$$\vec F=\nabla (4x^2+3y^4)=8xi+12y^3j$$

Use polar substitution for unit circle C: $x^2+y^2=1$, $$x=\cos\theta\implies dx=-\sin\theta d\theta\quad \text{&} \quad y=\sin\theta\implies dy=\cos\theta d\theta$$

$$\int_C\vec F\cdot d\vec r=\int_C(8xi+12y^3j)\cdot (dxi+dyj)$$$$=\int_C(8xdx+12y^3dy)$$ $$=\int_0^{\pi/2}(-8\cos\theta\sin\theta d\theta+12\sin^3\theta \cos\theta d\theta)$$ $$=-8\int_0^{\pi/2}\cos\theta\sin\theta d\theta+12\int_0^{\pi/2} \sin^3\theta \cos\theta d\theta$$

Answer 3

$$\vec F=\nabla (4x^2+3y^4)=8xi+12y^3j$$

From equation of unit circle $C$: $x^2+y^2=1\implies dy=\dfrac{-xdx}{y}$,

$$\int_C\vec F\cdot d\vec r=\int_C(8xi+12y^3j)\cdot (dxi+dyj)$$$$=\int_C(8xdx+12y^3dy)$$ $$=\int_1^0\left(8xdx+12y(1-x^2)\left(\frac{-xdx}{y}\right)\right)$$ $$=\int_1^0\left(-4x+12x^3\right)dx$$