Evaluar

Question

<blockquote> <p>$$\oint_{C}(y-\sin x)dx+\cos x dy$ $ triángulo: $$C=\{y=0,x=2\pi,\pi y=2x\}$ $</p> </blockquote> <p><img src="http://i.stack.imgur.com/miCLq.png" alt="enter image description here"></p> <p><strong>Mi intento:</strong></p> <p>Aplicando el teorema de Green</p> <p>$$\oint_{C}\underbrace{(y-\sin x)}_{P}dx+\underbrace{\cos x}_{Q} dy=\iint\bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\bigg)$$</p> <p>$$=\iint\bigg(-\sin x-1 \bigg)dxdy$$</p> <p>$$=\int_{0}^{4}\bigg[\int_{0}^{2 \pi}\bigg(-\sin x-1\bigg)dx\bigg]dy$$</p> <p>$$\int_{0}^{4}\bigg(\cos(2 \pi) -2 \pi\bigg)dy$$</p> <p>$$=\int_{0}^{4}\bigg(1-2 \pi\bigg)dy=\boxed{\color{red}{4-8\pi}}$$</p> <blockquote> <p>¿Es correcto, abrir área ser $ \color{blue}{4\pi}$?</p> </blockquote>

Answer 1

Dado

$$ \oint_C \Big( ( y - \sin(x) ) d x + \cos(x) d y \Big). $$

Usando el Verde del teorema, usted debe utilizar $C$, así que usted consigue

$$ \oint_C \Big( ( y - \sin(x) ) d x + \cos(x) d y \Big) = \oint_C \Big( P d x + Q d y \Big) = \int_0^{2\pi} d x \int_0^{2x/\pi} d y \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\\ = - \int_0^{2\pi} d x \int_0^{2x/\pi} d y \Big( \sin(x) + 1 \Big). $$

Esto trabaja

$$ - \int_0^{2\pi} d x \int_0^{2x/\pi} d y\Big( \sin(x) + 1 \Big) = - \frac{2}{\pi} \int_0^{2\pi} d x \Big( x \sin(x) + x \)\\ = \frac{2}{\pi} \left[ x \cos(x) - \sin(x) - \frac{1}{2} x^2 \right]_0^{2\pi} = 4 - 4\pi $$