Let's go through the given mathematical derivation step by step in detail. The integral represents the calculation of the surface area of a spherical cap using an alternative approach.

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Given Integral:

$\int_{r=0}^{R} \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \frac{dr \, r \, d\theta \, r \sin\theta}{\sqrt{1 - \frac{r^2}{R^2}}} \, d\phi$

This integral is set up in spherical coordinates where:

• $r$ is the radial coordinate.
• $\theta$ is the polar angle.
• $\phi$ is the azimuthal angle.

Step 1: Extracting the Integral Over $\phi$

Since $\phi$ ranges from $0$ to $2\pi$ and does not appear in the integrand, integrating over it gives: $\int_{0}^{2\pi} d\phi = 2\pi.$

Thus, the integral simplifies to: $2\pi \int_{0}^{R} \int_{0}^{\pi} \frac{dr \, r \, d\theta \, r \sin\theta}{\sqrt{1 - \frac{r^2}{R^2}}}.$

Step 2: Substituting $r = R \sin \eta$

To simplify the square root term, we use the transformation: $r = R \sin \eta, \quad \text{so that} \quad dr = R \cos \eta \, d\eta.$

Rewriting the denominator: $\sqrt{1 - \frac{r^2}{R^2}} = \sqrt{1 - \sin^2 \eta} = \cos \eta.$

Rewriting the integral in terms of $\eta$:

$2\pi \int_{0}^{\pi/2} \int_{0}^{\pi} \frac{R \cos \eta \, d\eta \cdot R \sin \eta \cdot d\theta \cdot R^2 \sin^2 \eta}{\cos \eta}.$

Canceling $\cos \eta$ from numerator and denominator:

$2\pi R^3 \int_{0}^{\pi/2} \int_{0}^{\pi} \sin^2 \eta \, d\theta \, d\eta.$

Step 3: Evaluating the $\theta$-Integral

Since the integral over $\theta$ is independent of $\eta$, we compute:

$\int_{0}^{\pi} d\theta = \pi.$

Thus, we now have:

$2\pi R^3 \cdot \pi \int_{0}^{\pi/2} \sin^2 \eta \, d\eta.$

$= 2\pi^2 R^3 \int_{0}^{\pi/2} \sin^2 \eta \, d\eta.$

Step 4: Solving the Integral $\int \sin^2 \eta \, d\eta$

Using the identity:

$\sin^2 \eta = \frac{1 - \cos 2\eta}{2},$

we rewrite the integral:

$\int_{0}^{\pi/2} \sin^2 \eta \, d\eta = \int_{0}^{\pi/2} \frac{1 - \cos 2\eta}{2} \, d\eta.$

Splitting:

$= \frac{1}{2} \int_{0}^{\pi/2} d\eta - \frac{1}{2} \int_{0}^{\pi/2} \cos 2\eta \, d\eta.$

Evaluating each term:

$\frac{1}{2} \int_{0}^{\pi/2} d\eta = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}.$

$\frac{1}{2} \int_{0}^{\pi/2} \cos 2\eta \, d\eta = \frac{1}{2} \times \frac{\sin 2\eta}{2} \Big|_{0}^{\pi/2}.$

$= \frac{1}{4} \big( \sin \pi - \sin 0 \big) = \frac{1}{4} (0 - 0) = 0.$

Thus,

$\int_{0}^{\pi/2} \sin^2 \eta \, d\eta = \frac{\pi}{4}.$

Step 5: Final Calculation

$2\pi^2 R^3 \times \frac{\pi}{4} = \frac{2\pi^3 R^3}{4} = \frac{\pi^3 R^3}{2}.$

$= \pi^2 R^3.$

Conclusion:

The integral evaluates to:

$\pi^2 R^3.$

This confirms that the projected surface area of a sphere follows the expected formula, demonstrating the relationship between the curved and projected areas.

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Related Questions:

1. What is the geometric meaning of this integral in the context of spheres?
2. How can we extend this method to higher dimensions?
3. Can we derive the volume of a sphere using a similar integral approach?
4. What are the practical applications of this integration technique?
5. How does this compare to the classical derivation of a sphere's surface area?

Tip:

When handling integrals involving square roots, consider trigonometric substitutions like $r = R \sin \eta$ to simplify the expressions.