The provided question is written in Japanese and involves multiple mathematical problems, including integrals, volume calculations, and region transformations. Let’s break it down step by step.

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Question Analysis:

1. (1) The task is to swap the order of integration for a double integral: $I = \int_0^1 \int_{-2\sqrt{1-y}}^{2\sqrt{1-y}} f(x, y) dx dy$

2. (2) Another integral $J$ is provided: $J = \frac{1}{2} \int_D (\tan x + y + | \tan x - y |) dx dy$ The region $D$ is defined as: $D = \{ (x, y) \,|\, 0 \leq x \leq \frac{\pi}{4}, \, 0 \leq y \leq 1 \}.$

3. (3) A 3D region $C$ is defined as: $x \leq 1 - y^2, \, x^2 - x + z^2 \leq 0$ The task is to calculate the volume $V(C)$.

4. (4) The final step asks to find the value of $S$, which likely corresponds to one of the calculated results above (to be clarified while solving).

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Solution:

Step 1: Swapping the order of integration

For $I = \int_0^1 \int_{-2\sqrt{1-y}}^{2\sqrt{1-y}} f(x, y) dx dy$, the integration region is described by: $-2\sqrt{1-y} \leq x \leq 2\sqrt{1-y}, \quad 0 \leq y \leq 1.$

To swap the order of integration, observe that:

• For fixed $x$, $y$ ranges between $0$ and $1 - \frac{x^2}{4}$ (since $x^2 \leq 4(1-y)$).

Thus, the swapped integral is: $I = \int_{-2}^2 \int_0^{1 - \frac{x^2}{4}} f(x, y) dy dx.$

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Step 2: Computing $J$

The integral $J$ involves the region $D = \{ (x, y) \,|\, 0 \leq x \leq \frac{\pi}{4}, \, 0 \leq y \leq 1 \}$: $J = \frac{1}{2} \int_0^{\frac{\pi}{4}} \int_0^1 (\tan x + y + | \tan x - y |) dy dx.$

To handle $| \tan x - y |$, split it into cases:

1. $\tan x \geq y$: $| \tan x - y | = \tan x - y$.
2. $\tan x < y$: $| \tan x - y | = y - \tan x$.

The integral becomes: $J = \frac{1}{2} \left[ \int_0^{\frac{\pi}{4}} \int_0^{\tan x} (2\tan x) dy dx + \int_0^{\frac{\pi}{4}} \int_{\tan x}^1 (2y) dy dx \right].$

Evaluate these step-by-step.

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Step 3: Volume of 3D region $C$

The region $C$ is defined as: $x \leq 1 - y^2, \quad x^2 - x + z^2 \leq 0.$

The second inequality, $x^2 - x + z^2 \leq 0$, describes a paraboloid that opens along the $x$-axis. Solving for $z^2$, we have: $z^2 \leq x - x^2.$

• For $z^2 \geq 0$, $x - x^2 \geq 0$, so $0 \leq x \leq 1$.
• The limits for $y$ are $-\sqrt{1-x} \leq y \leq \sqrt{1-x}$.

Thus, the volume is: $V(C) = \int_0^1 \int_{-\sqrt{1-x}}^{\sqrt{1-x}} \int_0^{\sqrt{x - x^2}} dz dy dx.$

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Step 4: Final Calculation for $S$

Once the above results are computed, use them to answer the final part.

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Next Steps:

Would you like detailed step-by-step calculations for one specific part (e.g., $J$, $V(C)$)? Let me know!

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Related Questions:

1. How do you determine the order of integration for a given double integral?
2. What are the geometric interpretations of the regions defined by the inequalities in $J$ and $V(C)$?
3. How do you handle absolute values within integrals effectively?
4. What techniques are useful for visualizing 3D regions like $C$?
5. How can symmetry simplify volume calculations?

Tip:

Always visualize the integration region (via graphs or sketches) to ensure the bounds are interpreted correctly!