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The given integral is:

\[
\int_0^2 r^2 \sqrt{4 - r^2} \, dr
\]

### Solution:

We can solve this by using a trigonometric substitution. Let’s use the substitution:

\[
r = 2 \sin \theta
\]

This implies:
\[
dr = 2 \cos \theta \, d\theta
\]
and
\[
r^2 = 4 \sin^2 \theta
\]

Substituting \( r = 2 \sin \theta \) into the integral, the limits of integration will change as follows:
- When \( r = 0 \), \( \theta = 0 \)
- When \( r = 2 \), \( \theta = \frac{\pi}{2} \)

Thus, the integral becomes:

\[
\int_0^{\frac{\pi}{2}} (4 \sin^2 \theta) \cdot (2 \cos \theta) \cdot 2 \cos \theta \, d\theta
\]

This simplifies to:

\[
16 \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \, d\theta
\]

We can now use the double-angle identity for \( \sin^2 \theta \) and \( \cos^2 \theta \):

\[
\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}
\quad \text{and} \quad
\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}
\]

Now substitute these identities into the integral:

\[
16 \int_0^{\frac{\pi}{2}} \frac{(1 - \cos(2\theta))(1 + \cos(2\theta))}{4} \, d\theta
\]

This simplifies to:

\[
4 \int_0^{\frac{\pi}{2}} (1 - \cos^2 (2\theta)) \, d\theta
\]

Now, simplify the expression:

\[
4 \left[ \int_0^{\frac{\pi}{2}} 1 \, d\theta - \int_0^{\frac{\pi}{2}} \cos^2 (2\theta) \, d\theta \right]
\]

The first integral is straightforward:

\[
\int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}
\]

For the second integral, we use the identity:

\[
\cos^2 (x) = \frac{1 + \cos (2x)}{2}
\]

Applying this, we get:

\[
\int_0^{\frac{\pi}{2}} \cos^2 (2\theta) \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left(1 + \cos (4\theta)\right) \, d\theta
\]

The first part is simple:

\[
\frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}
\]

The second part is:

\[
\frac{1}{2} \int_0^{\frac{\pi}{2}} \cos (4\theta) \, d\theta = 0
\]

because the integral of \( \cos (4\theta) \) over \( [0, \frac{\pi}{2}] \) is zero.

Thus, the second integral is \( \frac{\pi}{4} \), and the full expression becomes:

\[
4 \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = 4 \cdot \frac{\pi}{4} = \pi
\]

### Final Answer:
\[
\boxed{\pi}
\]

Do you want more details or have any questions?

Here are some relative questions to explore:
1. How does trigonometric substitution work in integrals?
2. Why is \( r = 2 \sin \theta \) a suitable substitution for this problem?
3. What is the double-angle identity and how is it useful?
4. What are some other methods for evaluating integrals involving square roots?
5. How does the integral of \( \cos(k\theta) \) behave over standard intervals?

**Tip:** When dealing with integrals involving square roots of quadratic expressions, trigonometric substitution often simplifies the expression significantly.

Evaluate the integral ∫ (from 0 to 2) r^2 * sqrt(4 - r^2) dr

This solution demonstrates how to evaluate the integral r^2 sqrt(4 - r^2) from 0 to 2 using trigonometric substitution. The step-by-step process includes the use of double angle identities and fundamental integral calculus techniques. Suitable for Calculus II students.

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