example video example evaluate ∮c(2y - esin(x))dx+(7x + √(y⁴ + 1))dy, where c is the circle x² + y² = 16…

example video example evaluate ∮c(2y - esin(x))dx+(7x + √(y⁴ + 1))dy, where c is the circle x² + y² = 16. solution the region d bounded by c is the disk x² + y² ≤ 16, so lets change to polar coordinates after applying greens theorem. ∮c(2y - esin(x))dx+(7x + √(y⁴ + 1))dy = ∬d∂/∂x(7x + √(y⁴ + 1)) - ∂/∂y(2y - esin(x))da = ∫₀²π∫₀⁴( )r dr dθ = 5∫₀²πdθ∫₀⁴r dr =

example video example evaluate ∮c(2y - esin(x))dx+(7x + √(y⁴ + 1))dy, where c is the circle x² + y² = 16. solution the region d bounded by c is the disk x² + y² ≤ 16, so lets change to polar coordinates after applying greens theorem. ∮c(2y - esin(x))dx+(7x + √(y⁴ + 1))dy = ∬d∂/∂x(7x + √(y⁴ + 1)) - ∂/∂y(2y - esin(x))da = ∫₀²π∫₀⁴( )r dr dθ = 5∫₀²πdθ∫₀⁴r dr =

Answer

Explanation:

Step1: Aplicar el teorema de Green

Según el teorema de Green, $\oint_{C}Pdx + Qdy=\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$, donde $P = 2y - e^{\sin(x)}$ y $Q=7x+\sqrt{y^{4}+1}$. Entonces, $\frac{\partial Q}{\partial x}=7$ y $\frac{\partial P}{\partial y}=2$. Entonces $\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA=\iint_{D}(7 - 2)dA=\iint_{D}5dA$.

Step2: Cambiar a coordenadas polares

En coordenadas polares, $x = r\cos\theta$, $y = r\sin\theta$ y $dA = rdr d\theta$. El disco $x^{2}+y^{2}\leq16$ se describe como $0\leq r\leq4$ y $0\leq\theta\leq2\pi$. Entonces $\iint_{D}5dA = 5\int_{0}^{2\pi}\int_{0}^{4}r drd\theta$.

Step3: Integrar con respecto a $r$

$\int_{0}^{4}r dr=\left[\frac{r^{2}}{2}\right]_{0}^{4}=\frac{4^{2}}{2}-\frac{0^{2}}{2}=8$.

Step4: Integrar con respecto a $\theta$

$5\int_{0}^{2\pi}d\theta\int_{0}^{4}r dr=5\int_{0}^{2\pi}d\theta\times8$. Y $\int_{0}^{2\pi}d\theta=\left[\theta\right]_{0}^{2\pi}=2\pi - 0 = 2\pi$. Entonces $5\times2\pi\times8 = 80\pi$.

Answer:

$80\pi$