Question

Let \( C \) be the boundary of the square with vertices \( (0,0), (1,0), (1,1), (0,1) \) oriented counterclockwise. Then the value of the line integral \[ \oint_C x^2y^2 dx + (x^2 - y^2) dy \] is ............ (rounded off to two decimal places).

Hint:

Use Green's theorem to simplify line integrals over closed curves into double integrals over the enclosed region.

Answer 1

Correct Answer: 0.65

Step 1: Apply Green's theorem.
\[ \oint_C P \, dx + Q \, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA, \] where \( P = x^2 y^2, \; Q = x^2 - y^2. \)

Step 2: Compute partial derivatives.
\[ \frac{\partial Q}{\partial x} = 2x, \frac{\partial P}{\partial y} = 2x^2 y. \] Hence, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x^2 y. \]

Step 3: Evaluate the double integral over the square \( 0 \le x, y \le 1. \)
\[ \iint_R (2x - 2x^2 y) \, dy \, dx = \int_0^1 \int_0^1 (2x - 2x^2 y) \, dy \, dx. \] Integrate w.r.t. \( y \): \[ = \int_0^1 [2x y - x^2 y^2]_0^1 dx = \int_0^1 (2x - x^2) dx. \] Integrate w.r.t. \( x \): \[ [ x^2 - \frac{x^3}{3} ]_0^1 = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.67. \]

Final Answer: \[ \boxed{0.67} \]