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calculate rda where r is the region bounded in the first quadrant by y 2 x 3 and the line y x

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Calculate ∬_RdA where R is the region bounded in the first quadrant by y^2=x^3 and the line y=x.

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Answer to a math question Calculate ∬_RdA where R is the region bounded in the first quadrant by y^2=x^3 and the line y=x.

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¡Absolutamente! Aquí se explica cómo calcular la integral doble y encontrar el área de la región. **1. Dibuja la región** Primero, siempre es una buena práctica visualizar la región de integración. * **y^2 = x^3:** Esta es una parábola lateral que se abre hacia la derecha. * **y = x:** Esta es una línea recta que pasa por el origen con una pendiente de 1. Se cruzan en el primer cuadrante formando la región R. **2. Determinar los límites de la integración** Dado que estamos tratando con una región algo inusual, es más fácil integrar primero con respecto a 'y' y luego a 'x' (dy dx). * **límites de y:** y va desde 0 hasta el punto donde se cruzan la línea y la curva. Para encontrar este punto, sustituye y=x en la ecuación y^2 = x^3. Esto nos da x^2 = x^3 => x = 1 (descartamos x = 0 ya que es el origen). Por tanto, 0 ≤ y ≤ 1. * **límites de x:** Para cada valor de y, x va desde la recta y=x hasta la curva y^2 = x^3. Resolviendo la ecuación de la curva para x, obtenemos x = y^(2/3). Entonces, y ≤ x ≤ y^(2/3). **3. Configurar la integral doble** La integral doble que representa el área es: ∬_RdA = ∫_(0)^(1) ∫_(y)^(y^(2/3)) dx dy **4. Evaluar la integral interna** ∫_(y)^(y^(2/3)) dx = [x]_(y)^(y^(2/3)) = y^(2/3) - y **5. Evaluar la integral exterior** ∫_(0)^(1) (y^(2/3) - y) dy = [3/5 * y^(5/3) - 1/2 * y^2 ]_(0)^(1 ) = 3/5 - 1/2 = 1/10 **6. El resultado** El valor de la integral doble es 1/10 unidades cuadradas. Esto representa el área de la región R.

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