d = {(r, θ)|1 ≤ r ≤ 2, 0 ≤ θ ≤ π}. therefore greens theorem gives the following. ∮_c 4y² dx + 10xy dy = ∬_d…

d = {(r, θ)|1 ≤ r ≤ 2, 0 ≤ θ ≤ π}. therefore greens theorem gives the following. ∮_c 4y² dx + 10xy dy = ∬_d ∂/∂x(10xy) - ∂/∂y(4y²) da = ∬_d ( ) da = 2 ∫_0^π ∫_1^2 (r sin(θ))r dr dθ = 2 ∫_0^π sin(θ) dθ ∫_1^2 r² dr = 2 ^π_0 r³/3_1^2 =

d = {(r, θ)|1 ≤ r ≤ 2, 0 ≤ θ ≤ π}. therefore greens theorem gives the following. ∮_c 4y² dx + 10xy dy = ∬_d ∂/∂x(10xy) - ∂/∂y(4y²) da = ∬_d ( ) da = 2 ∫_0^π ∫_1^2 (r sin(θ))r dr dθ = 2 ∫_0^π sin(θ) dθ ∫_1^2 r² dr = 2 ^π_0 r³/3_1^2 =

Answer

Explanation:

Step1: Calcular derivadas parciales

$$\frac{\partial}{\partial x}(10xy) = 10y$$, $$\frac{\partial}{\partial y}(4y^{2}) = 8y$$. Entonces, $$\frac{\partial}{\partial x}(10xy)-\frac{\partial}{\partial y}(4y^{2})=10y - 8y=2y$$.

Step2: Convertir a coordenadas polares

Sabemos que $$y = r\sin(\theta)$$ y $$dA = r\ dr\ d\theta$$ en coordenadas polares. El dominio $$D={(r,\theta)|1\leq r\leq2,0\leq\theta\leq\pi}$$. Así, la integral doble se convierte en $$2\int_{0}^{\pi}\int_{1}^{2}(r\sin(\theta))r\ dr\ d\theta$$.

Step3: Separar la integral doble en un producto de integrales

Usando la propiedad $$\int_{a}^{b}\int_{c}^{d}f(x)g(y)\ dx\ dy=\int_{a}^{b}f(x)\ dx\int_{c}^{d}g(y)\ dy$$, tenemos $$2\int_{0}^{\pi}\sin(\theta)\ d\theta\int_{1}^{2}r^{2}\ dr$$.

Step4: Calcular la integral con respecto a $$\theta$$

$$\int_{0}^{\pi}\sin(\theta)\ d\theta=-\cos(\theta)\big|_{0}^{\pi}=-( \cos(\pi)-\cos(0))=-(- 1 - 1)=2$$.

Step5: Calcular la integral con respecto a $$r$$

$$\int_{1}^{2}r^{2}\ dr=\frac{r^{3}}{3}\big|_{1}^{2}=\frac{2^{3}}{3}-\frac{1^{3}}{3}=\frac{8 - 1}{3}=\frac{7}{3}$$.

Step6: Calcular el resultado final

Multiplicamos los resultados de las integrales: $$2\times2\times\frac{7}{3}=\frac{28}{3}$$.

Answer:

$$\frac{28}{3}$$