To solve the given problem, we need to evaluate the line integral of the vector field $\vec{F}$ around the triangle $ABC$ in the $XY$-plane, where the vertices are $A(0,0)$, $B(2,0)$, and $C(2,1)$. The vector field is given as:

$\vec{F} = \big(2x + y^2\big)\hat{i} + \big(3y - 4x\big)\hat{j}.$

Strategy

We will apply Green's theorem, which simplifies the computation of line integrals in closed curves. Green's theorem states:

$\oint_{\partial R} \vec{F} \cdot d\vec{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA,$

where $\vec{F} = P(x, y)\hat{i} + Q(x, y)\hat{j}$.

Here:

• $P(x, y) = 2x + y^2$,
• $Q(x, y) = 3y - 4x$.

Step 1: Compute $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$

1. $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3y - 4x) = -4$,
2. $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x + y^2) = 2y$.

Thus: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -4 - 2y.$

Step 2: Set up the double integral over the triangular region

The vertices $A(0, 0)$, $B(2, 0)$, and $C(2, 1)$ form a triangular region $R$. The equations of the sides of the triangle are:

1. From $A(0, 0)$ to $B(2, 0)$: $y = 0$, $x$ ranges from 0 to 2.
2. From $B(2, 0)$ to $C(2, 1)$: $x = 2$, $y$ ranges from 0 to 1.
3. From $C(2, 1)$ to $A(0, 0)$: $y = \frac{1}{2}x$, $x$ ranges from 0 to 2.

The double integral will be evaluated over the region $R$, which is bounded by: [ x \in [0, 2], \quad y \in [0, \frac{x}{2}]. ]

Step 3: Evaluate the integral

$\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_R (-4 - 2y) \, dA.$

In terms of $x$ and $y$, the limits are:

• $x$ ranges from 0 to 2,
• $y$ ranges from 0 to $\frac{x}{2}$.

Thus: $\iint_R (-4 - 2y) \, dA = \int_0^2 \int_0^{x/2} (-4 - 2y) \, dy \, dx.$

1. Evaluate the inner integral with respect to $y$: $\int_0^{x/2} (-4 - 2y) \, dy = \left[ -4y - y^2 \right]_0^{x/2} = -4\left(\frac{x}{2}\right) - \left(\frac{x}{2}\right)^2 = -2x - \frac{x^2}{4}.$

2. Evaluate the outer integral with respect to $x$: $\int_0^2 \left(-2x - \frac{x^2}{4}\right) dx = \int_0^2 (-2x) \, dx + \int_0^2 \left(-\frac{x^2}{4}\right) dx.$

For the first term: $\int_0^2 -2x \, dx = \left[-x^2\right]_0^2 = -4.$

For the second term: $\int_0^2 -\frac{x^2}{4} \, dx = -\frac{1}{4} \int_0^2 x^2 \, dx = -\frac{1}{4} \left[\frac{x^3}{3}\right]_0^2 = -\frac{1}{4} \cdot \frac{8}{3} = -\frac{2}{3}.$

Combine the results: $\int_0^2 \left(-2x - \frac{x^2}{4}\right) dx = -4 - \frac{2}{3} = -\frac{14}{3}.$

Final Answer:

$\oint_{\partial R} \vec{F} \cdot d\vec{r} = -\frac{14}{3}.$

Let me know if you'd like more details or have any questions!

5 Related Questions:

1. What is Green's theorem, and how is it applied in general?
2. How can we parameterize the line integral without using Green's theorem?
3. What are the conditions for Green's theorem to hold?
4. How do we compute line integrals for non-conservative vector fields?
5. What are the applications of Green's theorem in physics and engineering?

Tip: When solving line integrals using Green's theorem, always verify that the curve is closed and oriented positively (counterclockwise).