Question

calcular a integral dupla de (x^2 + 4xy^3)dA limitada pela região y=x^2 e y=2x

Answer 1

(1) A região é definida asim: $R=\left\{(x,y): 0\leq x\leq 2~,~x^2\leq y\leq 2x\right\}$

(2) Nós calcularemos a siguente integral

                            $\displaystyle \iint\limits_R (x^2 + 4xy^3)dA$


$\displaystyle \iint\limits_R (x^2 + 4xy^3)dA=\int_{0}^{2}\int_{x^2}^{2x}(x^2 + 4xy^3)~dy~dx\\ \\ \\ \iint\limits_R (x^2 + 4xy^3)dA=\int_{0}^{2}\left.\left(x^2y+xy^4\right)\right|_{y=x^2}^{y=2x}~dx\\ \\ \\ \iint\limits_R (x^2 + 4xy^3)dA=\int_{0}^{2}(2x^3+16x^5)-(x^4-x^9)~dx\\ \\ \\ \iint\limits_R (x^2 + 4xy^3)dA=\int_{0}^{2}2x^3+16x^5-x^4-x^9~dx$


$\displaystyle \iint\limits_R (x^2 + 4xy^3)dA=\left.\left(\dfrac{x^4}{2}+\dfrac{8x^6}{3}-\dfrac{x^5}{5}-\dfrac{x^{10}}{10}\right)\right|_{0}^{2}\\ \\ \\ \iint\limits_R (x^2 + 4xy^3)dA=\dfrac{2^4}{2}+\dfrac{8\cdot2^6}{3}-\dfrac{2^5}{5}-\dfrac{2^{10}}{10}\\ \\ \\ \boxed{\boxed{\iint\limits_R (x^2 + 4xy^3)dA=\dfrac{1048}{15}}}$