Question

Evaluate the line integral [(2x-y+4)dx + (5y+3x-6)dy] around a circle of radius 4 with center at (0,0).

Answer 1

Evaluate the line integral [(2x-y+4)dx + (5y+3x-6)dy] around a circle of radius 4 with center at (0,0).

$$\int_{L} (2x - y + 4) \, dx + (5y + 3x - 6) \, dy$$

a) The equation of a circle of radius 4 with center at (0,0) in parametric form is

$$x = 2 \text{cost}$$ $$y = 2 \text{sint}$$ $$0 \leq t \leq 2\pi$$ $$dx = -2 \text{sint} \, dt$$ $$dy = 2 \text{cost} \, dt$$

Hence

$$\begin{aligned} \int_{L} (2x - y + 4) \, dx + (5y + 3x - 6) \, dy &= \\ &= \int_{0}^{2\pi} \left( - \left( 4 \text{cost} - 2 \text{sint} + 4 \right) 2 \text{sint} + \left( 10 \text{sint} + 6 \text{cost} - 6 \right) 2 \text{cost} \right) dt \\ &= \int_{0}^{2\pi} \left( 12 \text{costsint} + 4 \sin^2 x + 12 \cos^2 x - 8 \text{sint} - 12 \text{cost} \right) dt \end{aligned}$$

Using the double-angle formulae

$$\cos 2x = 2 \cos^2 x - 1$$ $$\begin{aligned} \int_{0}^{2\pi} \left( 12 \text{costsint} + 4 \cos 2t + 8 - 8 \text{sint} - 12 \text{cost} \right) dt \\ = \left( 6 \sin^2 t + 2 \sin 2t + 8t + 8 \cos t - 12 \sin t \right) \Big|_{0}^{2\pi} \\ = 6 \times 0 + 2 \times 0 + 16\pi + 8 \times 1 - 12 \times 0 - 6 \times 0 - 2 \times 0 - 0 - 8 \times 1 + 12 \times 0 = 16\pi \end{aligned}$$