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The given expression is a double integral in polar coordinates, typically used to compute the volume of a region. Here's the integral:

\[
V = \int_0^\pi \int_0^{2\sin(\theta)} \sqrt{4 - r^2} \, r \, dr \, d\theta
\]

To solve this, follow these steps:

1. **Integrate with respect to \( r \)** (the inner integral).
2. **Substitute limits for \( r \)**.
3. **Integrate with respect to \( \theta \)** (the outer integral).

### Step 1: Inner Integral
The inner integral is:

\[
\int_0^{2\sin(\theta)} \sqrt{4 - r^2} \, r \, dr
\]

Use substitution to simplify it by letting \( u = 4 - r^2 \), which gives \( du = -2r \, dr \).

### Step 2: Solve the outer integral
After evaluating the inner integral, substitute it into the outer integral and solve with respect to \( \theta \).

Would you like a detailed step-by-step solution for each of these steps? 

Also, here are 5 questions that build on this concept:
1. What is the purpose of using polar coordinates in volume integrals?
2. How would the limits of integration change if the shape or region described changes?
3. What substitutions are helpful in dealing with square root terms in integrals?
4. Can this integral be solved using numerical methods if it becomes too complex analytically?
5. How does the sine function in the upper limit affect the outcome?

### Tip:
When integrating using polar coordinates, always remember to include the extra factor of \( r \) to account for the Jacobian of the transformation from Cartesian to polar coordinates.

V = ∫(0 to π) ∫(0 to 2sin(θ)) sqrt(4 - r^2) * r dr dθ

Explore how to calculate volume using a double integral in polar coordinates. This problem focuses on integrating a region bounded by a trigonometric function and involves key calculus concepts such as polar coordinate transformation, substitution methods, and evaluating double integrals.

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