Suppose $F(x, y) = xyi + (x - y)$ and $C$ is the triangle with vertices $(-4,0)$, $(0,4)$, and $(4,0)$.

Question

Suppose $F(x, y) = xyi + (x - y)$ and $C$ is the triangle with vertices $(-4,0)$, $(0,4)$, and $(4,0)$. Let the line segment from $(4,0)$ to $(-4, 0)$ is $C_1$, the line segment from $(-4,0)$ to $(0, 4)$ is $C_2$, and the line segment from $(0,4)$ to $(4, 0)$ is $C_3$.

(a) Find the line integral of $F$ along each segment of the triangle.

(b) Find the circulation of $F$ around $C$.

(c) Find the circulation of $F$ around $-C$.

Parameterizations:

• For segment $C_1$ from $(4,0)$ to $(-4,0)$, I got $r(t) = -(8t)i$
• For segment $C_2$ from $(-4,0)$ to $(0,4)$, I got $r(t) = (4t)i + (4t)j$
• For segment $C_3$ from $(0,4)$ to $(4,0)$, I got $r(t) = (4t)i - (4t)j$

So basically what I've done for the segment from $(-4,0)$ to $(0,4)$ is this:

$F = (xy)i + (x - y)j $

$r(t) = (4t)i + (4t)j $

$F(r(t)) = [(4t)(4t)]i + [(4t) - (4t)]j = (16t^2)i $

$r'(t) = 4i + 4j $

$F(r(t))•r'(t) = (16t^2)i • (4i + 4j) = 64t^2 $

=> For $0 \le t \le 1 $

==> Integral from $0$ to $1$ of $64t^2 $

===> $(64/3)t^3$ evaluated from $0$ to $1 $

====> Integral $F•dr = 64/3 $

I don't understand where I'm going wrong... could someone point that out to me please?

Answer 1

Your parameterizations are incorrect. Putting $0$ and $1$ to your parameterization of $C_2$ yields respectively $(0,0)$ and $(4,4)$, which are not the endpoints of the segment.

The correct parameterizations (if the integrations are from $0$ to $1$) are

• $C_1\rightarrow r(t)=(4-8t)i$
• $C_2\rightarrow r(t)=(4t-4)i+(4t)j$
• $C_3\rightarrow r(t)=(4t)i+(4-4t)j$

Now let's try to obtain the line integral of $C_2$:

$F = (xy)i + (x - y)j $

$r(t) = (4t-4)i + (4t)j $

$F(r(t)) = [(4t-4)(4t)]i + [(4t-4) - (4t)]j = (16t^2-16t)i+(-4)j $.

Can you continue from there?