The integral shown in the image is:

$\int_0^r \sqrt{r^2 - x^2} \, dx$

This integral is the area of a quarter-circle with radius $r$. Let me solve it step-by-step.

Step 1: Recognizing the Geometry

The integrand $\sqrt{r^2 - x^2}$ represents the equation of a semicircle (upper half) in the Cartesian plane. Therefore, integrating from $x = 0$ to $x = r$ will give one-quarter of the area of a full circle.

The area of a full circle is $\pi r^2$, so the integral result is:

$\text{Result} = \frac{1}{4} \pi r^2$

Step 2: Solving Analytically

To confirm this result, we can evaluate the integral using a substitution method.

1. Substitute $x = r \sin\theta$, so $dx = r \cos\theta \, d\theta$.
2. Then, $r^2 - x^2 = r^2 (1 - \sin^2\theta) = r^2 \cos^2\theta$.
3. The bounds transform as follows:
   • When $x = 0$: $\theta = 0$,
   • When $x = r$: $\theta = \frac{\pi}{2}$.

The integral becomes: $\int_0^r \sqrt{r^2 - x^2} \, dx = \int_0^{\pi/2} \sqrt{r^2 \cos^2\theta} \cdot r \cos\theta \, d\theta$

Simplify: $\int_0^{\pi/2} r^2 \cos^2\theta \, d\theta$

4. Use the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$: $\int_0^{\pi/2} r^2 \cos^2\theta \, d\theta = r^2 \int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta$

Split the integral: $\int_0^{\pi/2} r^2 \cos^2\theta \, d\theta = \frac{r^2}{2} \int_0^{\pi/2} 1 \, d\theta + \frac{r^2}{2} \int_0^{\pi/2} \cos(2\theta) \, d\theta$

• The first term: $\frac{r^2}{2} \int_0^{\pi/2} 1 \, d\theta = \frac{r^2}{2} \cdot \frac{\pi}{2} = \frac{\pi r^2}{4}$

• The second term: $\frac{r^2}{2} \int_0^{\pi/2} \cos(2\theta) \, d\theta = \frac{r^2}{2} \cdot \left[ \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} = 0$

Thus: $\int_0^r \sqrt{r^2 - x^2} \, dx = \frac{\pi r^2}{4}$

Final Answer:

$\int_0^r \sqrt{r^2 - x^2} \, dx = \frac{\pi r^2}{4}$

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

Related Questions:

1. How do you compute the area of a full circle using integration?
2. How does the substitution $x = r\sin\theta$ simplify integrals of this type?
3. Can this problem be solved using polar coordinates directly?
4. What other physical problems result in quarter-circle integrals?
5. How does the geometric interpretation of integrals relate to trigonometric substitutions?

Tip:

For integrals involving $\sqrt{a^2 - x^2}$, trigonometric substitution ($x = a \sin\theta$) is often the best approach to simplify the problem systematically.