What is the line integral for $F=(y^2, x^2)$

Question

When C is the curve along the sides on the triangle with corners in $(0,0)$, $(1,0)$ and $(0,1)$ with counter-clockwise (positive) direction. Then $\int_0^.Fdr$ is? Do I have to make a parameterization of the triangle? How do i go about that? Thanks

Answer 1

Make it by pieces. For example, the side $\;(0,0) \to (1,0)\;$ ca be parametrized as

$$C_1:\;\;r(t)=(t,0)\;,\;\;t\in[0,1]\implies F(x(t),y(t))=(0^2,t^2)\;,\;\;r'(t)=(1,0)\implies$$

$$\int_{C_1} \vec F\cdot d\vec r=\int_0^1 F(r(t))\cdot r'(t)\,dt=\int_0^1 0\,dt=0$$

For the side $\;(1,0)\to(0,1)\;$ we have

$$C_2:\;\;r(t)=(1-t,t)\;,\;\;0\le t\le 1\implies F(r(t))=(t^2,(1-t)^2)\;,\;\;r'(t)=(-1,1)\implies$$

$$\int_{C_2}\vec F\cdot d\vec r=\int_0^1(t^2,(1-t)^2)\cdot(-1,1)\,dt=\int_0^1\left(-t^2+(1-t)^2\right)\,dt=$$

$$=\int_0^1(1-2t)\,dt=\left.(t-t^2)\right|_0^1=0$$

Now you try to do as above for the third and last side of the triangle.

Answer 2

Consider $\vec V=\langle \frac{x^3}{3}, \frac{y^3}{3} \rangle$. $\nabla V=F$. Let $C$ be any curve from $(0,0)$ to $(0,0)$. So by the fundamental theorem of calculus,

$$\int_{C} \nabla V \cdot \vec dr=?$$