Question

Let \( \Gamma \) be the positively oriented circle \( x^2 + y^2 = 9 \) in the xy-plane. If \[ \oint_{\Gamma} (3y + e^x \sin x) \, dx + \left( 7x + \sqrt{e^y + 2} \right) \, dy = \alpha \pi, \] where \( \alpha \) is a real constant, then \( \alpha \) is equal to ________.

Hint:

Use Green’s Theorem to convert a line integral around a closed curve into a double integral over the region enclosed by the curve.

Answer 1

Correct Answer: 36

This problem involves evaluating the line integral over a positively oriented circle. We can use Green’s Theorem to convert the line integral into a double integral. The line integral is given by: \[ \oint_{\Gamma} P(x, y) \, dx + Q(x, y) \, dy \] Where \( P(x, y) = 3y + e^x \sin x \) and \( Q(x, y) = 7x + \sqrt{e^y + 2} \). According to Green’s theorem: \[ \oint_{\Gamma} P(x, y) \, dx + Q(x, y) \, dy = \iint_{\text{region}} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] Now, compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = 7, \quad \frac{\partial P}{\partial y} = 3 \] So, the integrand becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 7 - 3 = 4 \] The region of integration is a circle of radius 3, so the area \( A \) of the region is: \[ A = \pi \times 3^2 = 9\pi \] Thus, the integral becomes: \[ \iint_{\text{region}} 4 \, dA = 4 \times 9\pi = 36\pi \] From the problem statement, this is equal to \( \alpha \pi \), so: \[ \alpha \pi = 36\pi \quad \Rightarrow \quad \alpha = 36 \] Thus, the value of \( \alpha \) is: 36