We are asked to evaluate the double integral

$I = \int\int_R \sqrt{\alpha^2 - x^2 - y^2} \, dx \, dy$

where the region $R$ is bounded by the semicircle $x^2 + y^2 = \alpha x$ in the first octant.

Step 1: Simplify the Region

First, rewrite the equation of the boundary $x^2 + y^2 = \alpha x$. This is a circle, but it can be made clearer by completing the square. Rewrite the equation as:

$x^2 - \alpha x + y^2 = 0$

Complete the square for $x$:

$(x - \frac{\alpha}{2})^2 + y^2 = \frac{\alpha^2}{4}$

This represents a circle with center $\left( \frac{\alpha}{2}, 0 \right)$ and radius $\frac{\alpha}{2}$. However, since we are in the first octant (where $x \geq 0$ and $y \geq 0$), we only consider the part of the circle in that region.

Step 2: Switch to Polar Coordinates

The integral can be simplified by switching to polar coordinates. In polar coordinates, we have:

$x = r \cos\theta, \quad y = r \sin\theta$

The equation of the circle $(x - \frac{\alpha}{2})^2 + y^2 = \frac{\alpha^2}{4}$ becomes $r = \alpha \cos\theta$, which gives the boundary of the region in polar form. The limits for $r$ will range from 0 to $\alpha \cos\theta$, and for $\theta$, from 0 to $\frac{\pi}{2}$ (since we are in the first octant).

Also, the area element $dx \, dy$ becomes $r \, dr \, d\theta$.

Step 3: Set Up the Integral

In polar coordinates, the expression $\sqrt{\alpha^2 - x^2 - y^2}$ becomes:

$\sqrt{\alpha^2 - r^2}$

Thus, the integral becomes:

$I = \int_0^{\frac{\pi}{2}} \int_0^{\alpha \cos\theta} \sqrt{\alpha^2 - r^2} \, r \, dr \, d\theta$

Step 4: Evaluate the Integral

First, evaluate the inner integral with respect to $r$:

$I_r = \int_0^{\alpha \cos\theta} \sqrt{\alpha^2 - r^2} \, r \, dr$

This is a standard integral that can be solved using the substitution $r = \alpha \sin u$, but another method involves the formula:

$\int_0^a r \sqrt{a^2 - r^2} \, dr = \frac{a^3}{3}$

Applying this formula, we get:

$I_r = \frac{(\alpha \cos\theta)^3}{3} = \frac{\alpha^3 \cos^3\theta}{3}$

Step 5: Integrate with Respect to $\theta$

Now, we integrate with respect to $\theta$:

$I = \int_0^{\frac{\pi}{2}} \frac{\alpha^3 \cos^3\theta}{3} \, d\theta$

Use the reduction formula for $\cos^3\theta$:

$\int_0^{\frac{\pi}{2}} \cos^3\theta \, d\theta = \frac{2}{3}$

Thus, the final result is:

$I = \frac{\alpha^3}{3} \cdot \frac{2}{3} = \frac{2\alpha^3}{9}$

Final Answer:

The value of the integral is:

$I = \frac{2\alpha^3}{9}$

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Would you like any further explanation on any of the steps? Here are some related questions:

1. How is the formula for integrating $\sqrt{\alpha^2 - r^2} \, r \, dr$ derived?
2. What is the geometric interpretation of the region bounded by $x^2 + y^2 = \alpha x$?
3. Why do we use polar coordinates in this problem?
4. How is the reduction formula for $\cos^3\theta$ derived?
5. Can you generalize this problem for different regions?

Tip: Polar coordinates are often useful when dealing with circular or radial symmetry in problems, as they simplify the limits and integrand.