Evaluate the double integral

*D*

*f*(*x*, *y*)*dA*

over the polar rectangular region *D*.

*f*(*x*, *y*) = arctan

*y*

*x*

, where *D* = (*r*, 𝜃)  1 ≤ *r* ≤ 3,

𝜋

4

≤ 𝜃 ≤

𝜋

3

To evaluate the double integral of \( f(x, y) = \arctan\left(\frac{y}{x}\right) \) over the polar region \( D \), we’ll proceed with the following steps:

1. **Rewrite \( f(x, y) \) in Polar Coordinates**:  
   In polar coordinates, we have \( x = r \cos \theta \) and \( y = r \sin \theta \). Therefore:
   \[
   \frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta.
   \]
   So, \( f(x, y) = \arctan\left(\frac{y}{x}\right) = \theta \).

2. **Set Up the Integral in Polar Coordinates**:  
   The region \( D \) is given by \( 1 \leq r \leq 3 \) and \( \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3} \). The area element \( dA \) in polar coordinates is \( r \, dr \, d\theta \).
   
   Therefore, the integral becomes:
   \[
   \iint_D f(x, y) \, dA = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \int_1^3 \theta \, r \, dr \, d\theta.
   \]

3. **Evaluate the Inner Integral with Respect to \( r \)**:  
   We first integrate with respect to \( r \), treating \( \theta \) as a constant:
   \[
   \int_1^3 \theta \, r \, dr = \theta \int_1^3 r \, dr.
   \]
   Calculating \( \int_1^3 r \, dr \):
   \[
   \int_1^3 r \, dr = \left. \frac{r^2}{2} \right|_1^3 = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4.
   \]
   Thus, the inner integral is:
   \[
   \int_1^3 \theta \, r \, dr = 4\theta.
   \]

4. **Evaluate the Outer Integral with Respect to \( \theta \)**:  
   Substitute \( 4\theta \) into the outer integral:
   \[
   \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 4\theta \, d\theta = 4 \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \theta \, d\theta.
   \]
   Calculating \( \int \theta \, d\theta \):
   \[
   \int \theta \, d\theta = \frac{\theta^2}{2}.
   \]
   So,
   \[
   \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \theta \, d\theta = \left. \frac{\theta^2}{2} \right|_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{\left(\frac{\pi}{3}\right)^2}{2} - \frac{\left(\frac{\pi}{4}\right)^2}{2} = \frac{\pi^2}{18} - \frac{\pi^2}{32}.
   \]
   To combine terms, find a common denominator (which is 288):
   \[
   \frac{\pi^2}{18} = \frac{16\pi^2}{288}, \quad \frac{\pi^2}{32} = \frac{9\pi^2}{288}.
   \]
   Thus,
   \[
   \frac{16\pi^2}{288} - \frac{9\pi^2}{288} = \frac{7\pi^2}{288}.
   \]

5. **Final Answer**:
   \[
   \iint_D f(x, y) \, dA = 4 \cdot \frac{7\pi^2}{288} = \frac{7\pi^2}{72}.
   \]

So, the value of the integral is:
\[
\frac{7\pi^2}{72}.
\]

Would you like a breakdown of any step or further details?

---

### Relative Questions
1. How do polar coordinates simplify integration in circular regions?
2. What are the advantages of converting functions to polar coordinates?
3. Can all integrals over circular regions be simplified using polar coordinates?
4. Why is \( dA = r \, dr \, d\theta \) in polar coordinates?
5. How does the arctangent function relate to the angle \( \theta \) in polar coordinates?

### Tip
In polar integration, always remember to include \( r \) in \( dA = r \, dr \, d\theta \) to account for the scaling in polar areas.

Evaluate the double integral ∫∫_D f(x, y) dA over the polar region D, where f(x, y) = arctan(y/x), D = (r, θ), 1 ≤ r ≤ 3, π/4 ≤ θ ≤ π/3.

Learn to evaluate a double integral ∫∫_D arctan(y/x) dA over a specified polar region using polar coordinates. This solution covers essential calculus concepts including double integrals and polar transformation, guiding through rewriting f(x, y) in terms of θ, setting up integrals with limits 1 ≤ r ≤ 3, π/4 ≤ θ ≤ π/3, and solving step-by-step.

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