0
$\begingroup$

$$\oint x\,dx\qquad C:\{x=0,y=0,y=-x+1 \} $$

My attempt:

$$\oint x\,dx=\int_{\uparrow}x\,dx+\int_{\nwarrow}x\,dx+\int_{\rightarrow}x\,dx$$

I don't know what should I do now, all the integral limits are from zero too one?

$\endgroup$
1
  • 1
    $\begingroup$ I understand what curve you mean to indicate, but the specification of $C$ isn't quite correct. $\endgroup$ May 17, 2016 at 12:25

4 Answers 4

1
$\begingroup$

The usual orientation is counter-clockwise so the path $C$, a triangle, would consist of the line segments from $(0,0)$ to $(1,0)$, then from $(1,0)$ to $(0,1)$ and finally from $(0,1)$ back to $(0,0)$.

If you really want to explicitly calculate the line integral, you should parametrize each part. With the given curves, that can be done very easily:

  • $(0,0)$ to $(1,0)$: take $x$ as parameter and $y=0$, then $x$ goes from $0$ to $1$;
  • $(1,0)$ to $(0,1)$: take $y$ as parameter and then $x=1-y$ with $y$ from $0$ to $1$;
  • $(0,1)$ to $(0,0)$: take $y$ as parameter and $x=0$, then $y$ goes from $1$ to $0$; or take $y=-t$ and then $t:0\to 1$.
$\endgroup$
1
$\begingroup$

The path you are given is closed, and the integrand has rotation zero, i.e it is the gradient of the scalar function $ F(x, y) = \left(\frac{x^2}{2}, 0\right) $. By the gradient theorem, the integral vanishes.

$\endgroup$
1
$\begingroup$

Hint Since the contour $C$ is closed, use Green's Theorem to rewrite the contour integral (which, as written, needs to be evaluated in three pieces) as a single (easy) double integral over the region $D$ enclosed by $C$:

$$\oint_C P \, dx + Q \,dy = \iint_D (Q_x - P_y) \,dx\,dy$$

$\endgroup$
2
  • $\begingroup$ I want to evalute using line integral, without Green's theorem $\endgroup$
    – Error 404
    May 17, 2016 at 12:23
  • $\begingroup$ You should specify that kind of requirement in the original question. Anyway, one can parametrize, e.g., the horizontal segment $H$ by $\gamma(t) = (t, 0)$, $t \in [0, 1]$, and so this part of the contour integral can be written as $\int_H x \,dx = \int_0^1 (t) d(t) = \left.\tfrac{1}{2} t^2 \right\vert_0^1 = \tfrac{1}{2}$. $\endgroup$ May 17, 2016 at 12:28
1
$\begingroup$

Extending StackTD answer, on the x-axis, from (0, 0) to (1, 0) we can take x itself as parameter with y identically 0. Since x goes form 0 to 1, the integral becomes $\int_0^1 xdx= \left[\frac{1}{2}x^2\right]_0^1= \frac{1}{2}$.

On the y-axis, from (0, 1) to (0, 0) we can take y as parameter with x identically 0. Since x is identically 0, dx= 0 and the integral, which contains no "y", is just 0.

On the line x+ y= 1 (same as y= -x+ 1 or x= -y+ 1) from (1, 0) to (0, 1) we can take either x or y as parameter. Using x as parameter we have again $\int xdx$ but now x is going from 1 down to 0 so we have $\int_1^0 xdx= -\int_0^1 dx= -\frac{1}{2}$. Using y as parameter, x= -y+ 1 and dx= -dy so $\int xdx= \int (-y+ 1)(-dy)= \int (y- 1)dy$. Going from (1, 0) to (0, 1), y goes from 0 up to 1 so the integral is $\int_0^1 (y- 1)dy= \left[\frac{1}{2}y^2- y\right]_0^1= \frac{1}{2}- 1= -\frac{1}{2}$. Either way, the integral on x+ y= 1 is -1/2 so the entire integral is 1/2+ 0- 1/2= 0.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .