Let \(\Delta\) be the triangular boundary in \(\mathbb{R}^2\) with vertices \((0,1)\), \((2,0)\), and \((2,1)\), traversed counterclockwise. Evaluate the line integral: \[I = \oint_{\Delta} y \, dx + 2x \, dy.\]
Solution
Goal: Evaluate the closed line integral \(\oint_\Delta y\,dx + 2x\,dy\) around the oriented triangle using Green’s Theorem.
1. Setting up Green’s Theorem:Proof:
1.1. Let \(D \subset \mathbb{R}^2\) be the filled triangular region enclosed by the positively oriented (counterclockwise) boundary curve \(\partial D = \Delta\).
1.2. The vector field components are:\[P(x, y) = y, \qquad Q(x, y) = 2x.\]
1.3. Both \(P\) and \(Q\) have continuous partial derivatives on \(\mathbb{R}^2\):\[\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x) = 2, \qquad \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y) = 1.\]
1.4. By Green’s Theorem:\[\oint_{\Delta} P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy = \iint_D (2 - 1) \, dA = \iint_D 1 \, dA = \operatorname{Area}(D).\]
2. Computation of the Area of the Triangular Region \(D\):Proof:
2.1. The vertices of the triangle are \(A = (0, 1)\), \(B = (2, 0)\), and \(C = (2, 1)\).
2.2. Notice that the side \(AC\) is the horizontal line segment from \((0, 1)\) to \((2, 1)\) of length \(2\).
2.3. The side \(BC\) is the vertical line segment from \((2, 0)\) to \((2, 1)\) of height \(1\).
2.4. Since \(AC\) is horizontal and \(BC\) is vertical, the angle at vertex \(C = (2, 1)\) is a right angle (\(90^\circ\)).
2.5. Thus \(D\) is a right-angled triangle with base \(b = 2\) and height \(h = 1\).
2.6. The area of \(D\) is:\[\operatorname{Area}(D) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 1 = 1.\]
3. Alternative Direct Line Integral Computation:Proof:
3.1. Path 1 (\(A(0,1) \to B(2,0)\)): Parameterize \(r_1(t) = (2t, 1 - t)\) for \(t \in [0, 1]\).- \(dx = 2\,dt\), \(dy = -dt\).- \(\int_{C_1} y\,dx + 2x\,dy = \int_0^1 (1 - t)(2) + 2(2t)(-1) \, dt = \int_0^1 (2 - 2t - 4t) \, dt = \int_0^1 (2 - 6t) \, dt = 2 - 3 = -1\).
3.2. Path 2 (\(B(2,0) \to C(2,1)\)): Parameterize \(r_2(t) = (2, t)\) for \(t \in [0, 1]\).- \(dx = 0\), \(dy = dt\).- \(\int_{C_2} y\,dx + 2x\,dy = \int_0^1 2(2) \, dt = 4\).
3.3. Path 3 (\(C(2,1) \to A(0,1)\)): Parameterize \(r_3(t) = (2 - 2t, 1)\) for \(t \in [0, 1]\).- \(dx = -2\,dt\), \(dy = 0\).- \(\int_{C_3} y\,dx + 2x\,dy = \int_0^1 (1)(-2) \, dt = -2\).
3.4. Summing all three edges:\[I = (-1) + 4 + (-2) = 1.\]
4. Conclusion:The value of the line integral is \(1\). Q.E.D.