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
Asked
Active
Viewed 45 times
0
-
Try splitting up the 3 different sides of the triangle into 3 lines and parametrise the lines – H_Hassan Mar 22 '17 at 12:29
2 Answers
2
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.
DonAntonio
- 211,718
- 17
- 136
- 287
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,
$$\int_{C} \nabla V \cdot \vec dr=?$$
Ahmed S. Attaalla
- 18,518
-
(1) First, you need a special form of the FTC for this two variables vector field. (2) The OP seems to be a beginner, learning how to parametrize and do line integrals by the very first definitions and methods. Conservative fields may be a little beyond his reach right now. – DonAntonio Mar 22 '17 at 17:32