I am trying to solve the following problem but I face difficulties with the results. Any suggestion?
The line integral $\int_C y^2 dx - x dy$, where $C$ is the boundary of the square $[-1,1]\times[-1,1]$ oriented counterclockwise, can be evaluated in two ways:
a) Using the definition of the line integral: To compute the line integral directly from its definition, we break the path $C$ into four segments corresponding to the sides of the square. For each segment, we parameterize the path, compute $dx$ and $dy$, substitute into the integral, and evaluate.
b) Using Green's theorem: Green's theorem relates a line integral around a simple closed curve $C$ to a double integral over the plane region $D$ bounded by $C$. It states that $\int_C P dx + Q dy = \int\int_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ where $P(x,y) = -x$ and $Q(x,y) = y^2$. We can compute the partial derivatives, substitute them into the double integral, and evaluate over the region $D$, which is the square \$[-1,1]\times[-1,1]$.
SOLUTION
a) To solve the line integral $\int_C y^2 dx - x dy$ using the definition, we need to consider the path $C$ which is the boundary of the square with vertices at $(-1,-1)$, $(-1,1)$, $(1,1)$, and $(1,-1)$, oriented counterclockwise. We'll break $C$ into four segments corresponding to the sides of the square:
$C_1$ from $(-1,-1)$ to $(-1,1)$ where $x = -1$ and y varies from $-1$ to $1$.
$C_2$ from $(-1,1)$ to $(1,1)$ where $y = 1$ and $x$ varies from $-1$ to $1$.
$C_3$ from $(1,1)$ to $(1,-1)$ where $x = 1$ and $y$ varies from $1$ to $-1$.
$C_4$ from $(1,-1)$ to $(-1,-1)$ where $y =-1$ and $x$ varies from $1$ to $-1$.
Now we'll compute the integral over each segment and sum them:
On $C_1$, dx = 0$ (since $x$ is constant), so the integral becomes: $$ \int_{-1}^{1} -x dy = \int_{-1}^{1} -(-1) dy = \int_{-1}^{1} dy = y\Big|_{-1}^{1} = 2 $$
On $C_2$, $dy = 0$ (since $y$ is constant), so the integral becomes: $$ \int_{-1}^{1} y^2 dx = \int_{-1}^{1} 1^2 dx = \int_{-1}^{1} dx = x\Big|_{-1}^{1} = 2 $$
On $C_3$, similar to $C_1$, $dx = 0$, so the integral becomes: $$ \int_{1}^{-1} -x dy = \int_{1}^{-1} -(1) dy = \int_{-1}^{1} dy = y\Big|_{-1}^{1} = 2 $$
On $C_4$, so the integral becomes: $$ \int_{1}^{-1} y^2 dx = \int_{1}^{-1} (-1)^2 dx = -\int_{-1}^{1} dx = -x\Big|_{-1}^{1} = -2 $$
Therefore, the total integral around $C$ is $2 + 2 + 2 -2 = 4$.
b)To correctly apply Green's Theorem for the line integral around the boundary $C $ of the square $[-1,1]\times[-1,1]$, we need to consider the functions $ P $ and $ Q $ as given in the integral:
[ \int_C y^2 dx - x dy ]
Here, $ P(x, y) = y^2 $ and $ Q(x, y) = -x $. According to Green's Theorem, the line integral over the closed curve $ C $ can be transformed into a double integral over the region $ R $ enclosed by $ C $:
$$ \oint_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $$
First, we calculate the partial derivatives of $ P$ and $ Q $:
- $$\frac{\partial Q}{\partial x} = \frac{\partial (-x)}{\partial x} = -1$$
- $$(\frac{\partial P}{\partial y} = \frac{\partial (y^2)}{\partial y} = 2y$$
So, the double integral becomes:
$$ \iint_R (-1 - 2y) dA $$
The region $ R $ is the square $[-1,1] \times [-1,1]$, thus the integral is over $ x $ from $-1$ to $1$ and $ y $ from $-1$ to $1$.
Calculating the double integral:
$$ \int_{-1}^{1} \int_{-1}^{1} (-1 - 2y) dx dy=-4$$