The integral given in the image is:

$\int_{a/4}^{a} \int_{0}^{\sqrt{2ax - x^2}} f(x, y) \, dy \, dx$

To reverse the order of integration (switching $dx$ and $dy$), we need to analyze the region of integration in the $xy$-plane and rewrite the limits accordingly.

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1. Analyze the Region of Integration

• The outer integral ($x$) runs from $x = a/4$ to $x = a$.
• The inner integral ($y$) runs from $y = 0$ to $y = \sqrt{2ax - x^2}$.

The curve $y = \sqrt{2ax - x^2}$ corresponds to the top boundary of the region, which is a portion of a downward-opening parabola. Rewriting it in terms of $x$:

$y^2 = 2ax - x^2 \quad \implies \quad x^2 - 2ax + y^2 = 0 \quad \implies \quad x = a \pm \sqrt{a^2 - y^2}.$

Since $x$ runs between $a/4$ and $a$, we focus on the right-hand branch:

$x = a - \sqrt{a^2 - y^2}.$

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2. New Limits for Reversed Integration

• $y$ now becomes the outer integral, running from $y = 0$ to $y = \sqrt{a}$ (the maximum height of the parabola).
• For a fixed $y$, $x$ runs from $x = a/4$ (the leftmost boundary) to $x = a - \sqrt{a^2 - y^2}$ (the parabola boundary).

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3. Reversed Integral

The integral with reversed order of integration is:

$\int_{0}^{\sqrt{a}} \int_{a/4}^{a - \sqrt{a^2 - y^2}} f(x, y) \, dx \, dy$

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Let me know if you need further clarification or detailed steps!

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Related Questions:

1. What does reversing the order of integration mean geometrically?
2. How would the region change if $f(x, y)$ were split into separate parts?
3. How do we determine the limits for more complex regions?
4. Can we solve the integral explicitly if $f(x, y) = 1$?
5. What is the significance of a parabola in such problems?

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Tip:

Visualizing the integration region with a sketch in the $xy$-plane greatly simplifies finding the new limits.