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To evaluate the given double integral, let's go through the problem step-by-step.

The integral is written in polar coordinates as:

\[
\int_0^{\pi} \int_0^5 r \cos(\theta) \, r \, dr \, d\theta
\]

The function \( f(r, \theta) = r \cos(\theta) \) is multiplied by \( r \) due to the Jacobian determinant when converting to polar coordinates.

### Step 1: Simplify the Integrand
The integrand becomes:
\[
r \cdot r \cos(\theta) = r^2 \cos(\theta)
\]

Thus, the integral can be rewritten as:
\[
\int_0^{\pi} \int_0^5 r^2 \cos(\theta) \, dr \, d\theta
\]

### Step 2: Integrate with Respect to \( r \)
Now, we focus on the inner integral with respect to \( r \):
\[
\int_0^{\pi} \left( \int_0^5 r^2 \cos(\theta) \, dr \right) d\theta
\]

Since \( \cos(\theta) \) is independent of \( r \), we can factor it out:
\[
= \int_0^{\pi} \cos(\theta) \left( \int_0^5 r^2 \, dr \right) d\theta
\]

Evaluate the inner integral:
\[
\int_0^5 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^5 = \frac{5^3}{3} = \frac{125}{3}
\]

Thus, the integral becomes:
\[
= \int_0^{\pi} \cos(\theta) \cdot \frac{125}{3} \, d\theta = \frac{125}{3} \int_0^{\pi} \cos(\theta) \, d\theta
\]

### Step 3: Integrate with Respect to \( \theta \)
Now, evaluate the remaining integral:
\[
\int_0^{\pi} \cos(\theta) \, d\theta
\]

The integral of \( \cos(\theta) \) is \( \sin(\theta) \):
\[
\int_0^{\pi} \cos(\theta) \, d\theta = \left[ \sin(\theta) \right]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0
\]

### Final Answer
Since this last integral evaluates to zero, the value of the entire double integral is:
\[
\int_0^{\pi} \int_0^5 r^2 \cos(\theta) \, dr \, d\theta = 0
\]

### Region \( R \)
The region \( R \) in polar coordinates is defined by:
- \( r \) ranging from 0 to 5 (a radius up to 5 units)
- \( \theta \) ranging from 0 to \( \pi \) (a half-circle in the first and second quadrants)

So, \( R \) represents the upper half of a disk with radius 5.

Would you like a more detailed breakdown of any part of this solution, or do you have questions about the region sketch?

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**Related Questions:**

1. What is the significance of using polar coordinates in double integrals?
2. How would the integral change if the limits for \( \theta \) were from \( 0 \) to \( 2\pi \)?
3. What is the purpose of the Jacobian in coordinate transformations?
4. How do we determine the appropriate region \( R \) in polar coordinates?
5. Can we convert any double integral from Cartesian to polar coordinates?

**Tip:** When integrating in polar coordinates, always remember to include \( r \) in the integrand to account for the area element \( dA = r \, dr \, d\theta \).

Evaluate the double integral ∫∫_R f(r, θ) dA with f(r, θ) = r cos(θ), over the region R defined by r from 0 to 5 and θ from 0 to π.

Step-by-step solution to evaluate the double integral of r cos(θ) over the region defined in polar coordinates with r from 0 to 5 and θ from 0 to π. Learn about the integration process in polar coordinates, including the Jacobian determinant. Suitable for undergraduate calculus students.

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