To evaluate the double integral \( \iint_{D}\left(x^{2}+6y\right) dA \), where \( D \) is bounded by \( y=x \), \( y=x^{3} \), and \( x \geq 0 \), we'll set up the integral limits based on the given bounds:
First, find the intersection points of \( y=x \) and \( y=x^{3} \):
\( x = x^{3} \) gives \( x = 0 \) or \( x = 1 \).
So, the limits of integration for \( x \) will be \( 0 \leq x \leq 1 \).
For \( y \), the lower bound is \( y = x \), and the upper bound is \( y = x^{3} \).
The integral becomes:
\[ \int_{0}^{1} \int_{x}^{x^{3}} \left(x^{2} + 6y\right) dy dx \]
Now integrate with respect to \( y \) first:
\[ \int_{0}^{1} \left[ x^{2}y + 3y^{2} \right]_{x}^{x^{3}} dx \]
\[ = \int_{0}^{1} \left( x^{5} + 3x^{6} - x^{3} - 3x^{2} \right) dx \]
Now integrate with respect to \( x \):
\[ \left[ \frac{1}{6}x^{6} + \frac{3}{7}x^{7} - \frac{1}{4}x^{4} - x^{3} \right]_{0}^{1} \]
\[ = \left( \frac{1}{6} + \frac{3}{7} - \frac{1}{4} - 1 \right) - 0 \]
\[ = \frac{7}{42} \]
So, the value of the double integral is \( \frac{7}{42} \) or approximately \( 0.1667 \).