Answer
$\dfrac{3}{10}$
Work Step by Step
The iterated integral can be calculated as:
$\ Area =\iint_{D} f(x,y) d A=\int_0^1 \int_{0}^{x^2} \ dy \ dx\\=\int_0^1 [x^2-0] \ dx \\=\int_0^1 x^2 \ dx \\=[\dfrac{x^3}{3}]_0^1 \\=\dfrac{1}{3}$
The average co-ordinate of $y$ is:
$\overline{f}=\dfrac{1}{Area} \iint_{D} f(x,y) d A\\=\dfrac{1}{(1/3)}\int_0^1 \int_{0}^{x^2} y \ dy \ dx\\=3 \int_0^1 \dfrac{x^4}{2}\\=(3) \times [\dfrac{x^5}{10}]_0^1 \\=(3) \times [\dfrac{(1)^5}{10}]\\=\dfrac{3}{10}$
