This is a calculus problem involving changing the order of integration in a double integral. Let's carefully analyze the given integral and steps for changing the order.

The integral is:

$I = \int_0^{\frac{1}{2}} \int_1^{1 + \sqrt{2y - y^2}} f(x, y) \, dx \, dy + \int_{\frac{1}{2}}^1 \int_{2y}^{1 + \sqrt{2y - y^2}} f(x, y) \, dx \, dy.$

Step 1: Understand the region of integration

The region is defined by the bounds:

1. For $0 \leq y \leq \frac{1}{2}$, $x$ varies from $x = 1$ to $x = 1 + \sqrt{2y - y^2}$.
2. For $\frac{1}{2} \leq y \leq 1$, $x$ varies from $x = 2y$ to $x = 1 + \sqrt{2y - y^2}$.

These describe two regions in the $x$-$y$ plane. Combining these, the overall region in the $x$-$y$ plane needs to be analyzed for converting the order of integration.

Step 2: Analyze boundaries

The boundaries are derived from the equations $x = 1$, $x = 1 + \sqrt{2y - y^2}$, and $x = 2y$, as well as the range for $y$: $0 \leq y \leq 1$.

• From $x = 1 + \sqrt{2y - y^2}$, rewrite as: $\sqrt{2y - y^2} = x - 1 \implies 2y - y^2 = (x - 1)^2.$ Expanding and rearranging gives: $y^2 - 2y + (x - 1)^2 = 0,$ which is a quadratic equation in $y$. This represents part of a parabolic boundary.

• The line $x = 2y$ is another boundary in the region.

Step 3: New bounds for $x$ and $y$

We need to rewrite the region with the roles of $x$ and $y$ swapped. Observing the graph, the new bounds are:

1. $x$ ranges from $1$ to $2$.
2. For a fixed $x$, $y$ ranges between the lines and curves that bound the region.

The new integral becomes: $I = \int_1^2 \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx,$ where $g_1(x)$ and $g_2(x)$ represent the lower and upper bounds for $y$ as a function of $x$.

• For $x \in [1, \frac{3}{2}]$, $y$ varies from $y = 0$ (bottom edge) to $y = \frac{x-1}{2}$ (from $x = 2y$).
• For $x \in [\frac{3}{2}, 2]$, $y$ varies from $y = 0$ to $y = 1 - \sqrt{2 - 2x + x^2}$ (from the parabola).

Final Answer

The changed order of integration is: $I = \int_1^{\frac{3}{2}} \int_0^{\frac{x-1}{2}} f(x, y) \, dy \, dx + \int_{\frac{3}{2}}^2 \int_0^{1 - \sqrt{2 - 2x + x^2}} f(x, y) \, dy \, dx.$

Would you like a detailed explanation of any step? Let me know!

Related Questions:

1. How do you derive the equations of boundaries in a double integral problem?
2. What are the general principles for changing the order of integration in double integrals?
3. How can the quadratic equation $y^2 - 2y + (x-1)^2 = 0$ be solved explicitly for $y$?
4. How does the geometric interpretation of integration regions simplify problems like this?
5. What methods can verify if the rewritten integral is equivalent to the original?

Tip:

When swapping the order of integration, always sketch the region to visualize the boundaries. This prevents errors in setting the new bounds!