Question

Let \( \Gamma \) denote the boundary of the square region \( R \) with vertices \( (0,0), (2,0), (2,2), (0,2) \), oriented in the counter-clockwise direction. Then \( \int_{\Gamma} (1 - y^2) dx + x \, dy = \underline{\hspace{1cm}}. \)

Hint:

Use Green's Theorem to convert line integrals into double integrals, simplifying the calculation of complex integrals over closed curves.

Answer 1

Correct Answer: 12

We apply Green's Theorem to solve this line integral. Green's Theorem states: \[ \int_{\Gamma} P \, dx + Q \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \] Here, \( P(x, y) = 1 - y^2 \) and \( Q(x, y) = x \). We compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = 1 \text{and} \frac{\partial P}{\partial y} = -2y. \] Thus, the double integral becomes: \[ \iint_R (1 + 2y) \, dA. \] Evaluating the integral over the square region, we get the value of the integral: \[ \boxed{12}. \]