The problem presented is to find the maximum value bounded by the curve $y = 2\sqrt{x}$, the vertical line $x = 9$, and the horizontal line $y = 0$, with a given density function $\rho(x, y) = x + 1$. Let's break down the process to solve it.

1. Step-by-step outline:

1.1. Area Bounded by the Curve:

The given curve is $y = 2\sqrt{x}$.

• This equation can be squared to express $x$ in terms of $y$: $y^2 = 4x \quad \Rightarrow \quad x = \frac{y^2}{4}$
• The region is bounded by the curve from $x = 0$ to $x = 9$, and by $y = 0$ (horizontal axis) up to the point where $y = 2\sqrt{9} = 6$.

1.2. Volume or Mass Calculation:

The mass can be calculated using the given density function $\rho(x, y) = x + 1$. The formula for the mass $M$ is the double integral of the density function over the region $R$:

$M = \iint_R \rho(x, y) \, dA = \int_0^6 \int_0^{\frac{y^2}{4}} (x + 1) \, dx \, dy$

1.3. Setting up the Integral:

The region of integration is described by:

• $y$ going from 0 to 6
• $x$ going from 0 to $\frac{y^2}{4}$

We now compute the inner integral first: