This is a problem requiring you to change the order of integration in a double integral. Let's work through it step by step:

Problem:

We are given the integral:

$\int_0^4 \int_{\frac{\sqrt{16 - x^2}}{8}}^{\sqrt{16 - x^2}} f(x, y) \, d\!y \, d\!x$

Our goal is to change the order of integration.

Step 1: Understand the integration bounds

The given bounds describe the region of integration. Let's analyze the limits for $y$ and $x$:

1. $x \in [0, 4]$: The outer integral tells us that $x$ ranges from 0 to 4.

2. For a fixed $x$, $y$ ranges from: $y \in \left[\frac{\sqrt{16 - x^2}}{8}, \sqrt{16 - x^2}\right]$ So, for each $x$, $y$ is bounded by two functions of $x$.

Step 2: Find the region of integration in the $xy$-plane

The integration limits describe the region in the $xy$-plane. To rewrite the integral with reversed order of integration, we need to:

1. Determine the full region of integration.
2. Express $x$ as a function of $y$ instead of $y$ as a function of $x$.

a) Analyze the condition $y \in \left[\frac{\sqrt{16 - x^2}}{8}, \sqrt{16 - x^2}\right]$:

• The lower bound for $y$ is $y = \frac{\sqrt{16 - x^2}}{8}$.
• The upper bound for $y$ is $y = \sqrt{16 - x^2}$.

These bounds indicate that for a fixed $x$, $y$ lies between these two curves.

b) Analyze the condition $x \in [0, 4]$:

• The variable $x$ ranges from 0 to 4.

c) Rewrite $y = \sqrt{16 - x^2}$ and $y = \frac{\sqrt{16 - x^2}}{8}$:

• $y = \sqrt{16 - x^2}$ corresponds to the top half of a circle with radius 4 and center at $(0, 0)$.
• $y = \frac{\sqrt{16 - x^2}}{8}$ is a scaled-down version of this curve.

The integration region lies between these two curves, for $x \geq 0$.

Step 3: Reverse the order of integration

To reverse the order of integration, we express $x$ as a function of $y$:

1. From $y = \sqrt{16 - x^2}$, solve for $x$: $x = \sqrt{16 - y^2}$
2. From $y = \frac{\sqrt{16 - x^2}}{8}$, solve for $x$: $x = \sqrt{16 - (8y)^2}$

Now, the bounds for $x$ become $x \in [\sqrt{16 - (8y)^2}, \sqrt{16 - y^2}]$, and $y$ ranges from: $y \in [0, \frac{1}{2}].$

Step 4: Write the integral with reversed bounds

The new order of integration is:

$\int_0^{\frac{1}{2}} \int_{\sqrt{16 - (8y)^2}}^{\sqrt{16 - y^2}} f(x, y) \, d\!x \, d\!y$

Final Answer:

The integral with reversed order of integration is:

$\int_0^{\frac{1}{2}} \int_{\sqrt{16 - (8y)^2}}^{\sqrt{16 - y^2}} f(x, y) \, d\!x \, d\!y$

Would you like further details or clarification?

Related Questions:

1. How do you verify the limits when changing the order of integration?
2. What is the geometric interpretation of the bounds in this problem?
3. How do scaling factors (like $8y$) affect the region of integration?
4. How do you handle similar problems for polar coordinates?
5. What are the common mistakes to avoid when reversing integration order?

Tip:

Always sketch the region of integration in the $xy$-plane to understand the bounds clearly before attempting to change the order of integration.