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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:
\[
\int_0^{\frac{y^2}{4}} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_0^{\frac{y^2}{4}}
= \frac{\left(\frac{y^2}{4}\right)^2}{2} + \frac{y^2}{4} 
= \frac{y^4}{32} + \frac{y^2}{4}
\]

#### **1.4. Outer Integral:**
Now integrate this result with respect to \( y \):
\[
M = \int_0^6 \left( \frac{y^4}{32} + \frac{y^2}{4} \right) \, dy
\]

Let's solve these two integrals separately:
- For \( \int_0^6 \frac{y^4}{32} \, dy \):
  \[
  \int_0^6 \frac{y^4}{32} \, dy = \frac{1}{32} \cdot \frac{y^5}{5} \Bigg|_0^6 = \frac{1}{32} \cdot \frac{6^5}{5} = \frac{7776}{160} = 48.6
  \]
- For \( \int_0^6 \frac{y^2}{4} \, dy \):
  \[
  \int_0^6 \frac{y^2}{4} \, dy = \frac{1}{4} \cdot \frac{y^3}{3} \Bigg|_0^6 = \frac{1}{4} \cdot \frac{6^3}{3} = \frac{216}{12} = 18
  \]

#### **1.5. Total Mass:**
Now, adding these two results together gives the total mass \( M \):
\[
M = 48.6 + 18 = 66.6
\]

### Final Answer:
The total mass of the region bounded by \( y = 2\sqrt{x} \), \( x = 9 \), and \( y = 0 \) with the density function \( \rho(x, y) = x + 1 \) is **66.6 units**.

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Do you want a deeper breakdown of any steps or have any questions?

### Related Questions:
1. How would the result change if the density function was different?
2. Can we interpret this result in terms of physics (e.g., center of mass)?
3. What is the significance of using a density function in this context?
4. How do we find the centroid of this bounded region?
5. Can we use different coordinates (polar, etc.) to solve similar problems?

#### Tip:
Always visualize the region of integration clearly to avoid errors in setting up the limits.

Find the mass bounded by y = 2√x, x = 9 and y = 0 with density function ρ(x, y) = x + 1.

Learn how to calculate the mass of the region bounded by the curve y = 2√x, the vertical line x = 9, and the horizontal line y = 0. The mass is calculated using the density function ρ(x, y) = x + 1 and involves setting up a double integral for precise evaluation. A step-by-step solution for understanding definite integrals in calculus.

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