Answer
$\dfrac{23}{84}$
Work Step by Step
Our aim is to calculate the area of the given surface.
The domain $D$ in Type-1 can be expressed as follows:
$D=\left\{ (x, y) | x^3 \leq y \leq x , \ 0 \leq x \leq 1 \right\}
$
Thus, $\iint_{D} (x^2+2y) \ dA=\int_{0}^{1} \int_{x^3}^{x} (x^2+2y) \ dy \ dx \\ =\int_0^1 [x^2y+y^2]_{x^3}^{x} \ dx \\= \int_0^1 [x^2 (x)+x^2-x^2 \times x^3 -(x^3)^2] d x \\ = [\dfrac{x^4}{4} ]_0^1+[\dfrac{x^3}{3} ]_0^1-[\dfrac{x^6}{6}]_0^1-[\dfrac{x^7}{7}]_0^1 \\= \dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{7} \\=\dfrac{23}{84}$
