We are given the double integral:

$\int_0^{81} \int_0^{\frac{y}{9}} \sqrt{y} \, dx \, dy$

Our goal is to reverse the order of integration. To do this, we first need to understand the region of integration in the $xy$-plane.

Step 1: Identify the limits of integration

• The outer integral has $y$ going from 0 to 81, so $y \in [0, 81]$.
• The inner integral has $x$ going from 0 to $\frac{y}{9}$, so $x \in [0, \frac{y}{9}]$.

Step 2: Determine the region of integration

• The variable $x$ is bounded by $x = 0$ and $x = \frac{y}{9}$.
• The variable $y$ is bounded between 0 and 81.

We can express the region in the $xy$-plane by rearranging the condition $x \leq \frac{y}{9}$, or equivalently:

$y \geq 9x$

Thus, the region is bounded by:

• $x$ going from 0 to 9 (since when $x = 9$, $y = 81$),
• For each $x$, $y$ ranges from $9x$ to 81.

Step 3: Reverse the order of integration

We can now write the integral with reversed order:

$\int_0^9 \int_{9x}^{81} \sqrt{y} \, dy \, dx$

This is the integral with the reversed order of integration.

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Would you like to proceed with evaluating the integral, or do you have any questions about this process?

Related Questions:

1. How do we evaluate a double integral?
2. What is the geometric interpretation of reversing the order of integration?
3. How do you handle integrals with non-rectangular regions?
4. What is the importance of the Jacobian when changing variables in multiple integrals?
5. Can the integral be simplified by using a substitution?

Tip: To better visualize the region of integration, sketching the graph of the equations $x = \frac{y}{9}$ and $y = 81$ can help clarify the bounds for $x$ and $y$.