I need some advice. The problem is to Evaluate the line integral by two methods $(a)$ directly and $(b)$ using Green's Theorem $$\oint_c y^2 \, d x + x^2 y \, dy $$ $C $ is the rectangle with vertices $\left( 0,0\right)$ , $\left( 5,0\right)$ , $\left( 5,4\right)$ and $\left( 0,4\right)$ So at first I solved using Green's Theorem $$ \int_0^ <4>\int_^ 2 x y - 2 y \, d x \, d y = 120$$ After that I tried to solve this problem using vector equations. $$r_1(t) = \Rightarrow \, dx = 5 \, dt \text < and >\, dy = 0 \,dt$$ $$r_2(t) = \Rightarrow \, dx = 0 \, dt \text < and >\, dy = 4 \,dt$$ $$r_3(t) = \Rightarrow \, dx = -5 \, dt \text < and >\, dy = 0 \,dt $$ $$r_4(t) = \Rightarrow \, dx = 0 \, dt \text < and >\, dy = -4 \,dt$$ And then just calculated $4$ integrals and added them. And after calculations I am not getting $120$ . So can you tell me if the idea is correct or not?
asked Nov 11, 2021 at 11:27 383 2 2 silver badges 10 10 bronze badges$ \displaystyle \oint_c y^2 \, d x + x^2 y \, dy$
$C$ is the rectangle with vertices $(0,0) , (5,0) , (5,4)$ and $(0,4)$ .
The line integral from $(0,0)$ to $(5, 0)$ is zero as $y = 0$ .
$r_2(t) = (5, 4t), 0 \leq t \leq 1, r_2'(t) = (0, 4)$
$ \displaystyle \int_0^1 5^2 \cdot 4t \cdot 4 ~ dt = 200$
$r_3(t) = (5-5t, 4), 0 \leq t \leq 1, r_3'(t) = ( -5, 0)$
$ \displaystyle \int_0^1 4^2 \cdot - 5 ~ dt = - 80$
The line integral from $(0, 4)$ to $(0, 0)$ is zero as $x = 0$ and $dx = 0$ .