https://file.aiquickdraw.com/mathbot/file/1730147932119hbr559a4.jpg

To solve this integral, we will change to polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \). The given region \( R \) is the first quadrant of the circle \( x^2 + y^2 = 4 \), which translates to \( r^2 = 4 \) or \( r = 2 \) in polar coordinates.

### Steps to Evaluate the Integral:

1. **Rewrite the Integral in Polar Coordinates:**

   In polar coordinates:
   - \( dA = r \, dr \, d\theta \),
   - \( x = r \cos \theta \),
   - \( y = r \sin \theta \).
   
   So, the integrand \( y e^x \) becomes:
   \[
   y e^x = (r \sin \theta) e^{r \cos \theta}.
   \]

   Therefore, the integral becomes:
   \[
   \int \int_R y e^x \, dA = \int_0^{\pi/2} \int_0^2 (r \sin \theta) e^{r \cos \theta} \, r \, dr \, d\theta.
   \]

2. **Set Up the Limits of Integration:**

   - \( r \) ranges from \( 0 \) to \( 2 \) (as the circle's radius is \( 2 \)).
   - \( \theta \) ranges from \( 0 \) to \( \pi/2 \) (as we are only in the first quadrant).

   Thus, we have:
   \[
   \int_0^{\pi/2} \int_0^2 r^2 \sin \theta \, e^{r \cos \theta} \, dr \, d\theta.
   \]

3. **Separate the Integral:**

   Since \( \sin \theta \) and \( e^{r \cos \theta} \) do not depend on \( r \) and \( \theta \) respectively, we can separate the integral as:
   \[
   \int_0^{\pi/2} \sin \theta \, d\theta \int_0^2 r^2 e^{r \cos \theta} \, dr.
   \]

4. **Evaluate Each Integral Separately:**

   1. **Integrate with respect to \( \theta \):**
      \[
      \int_0^{\pi/2} \sin \theta \, d\theta = [-\cos \theta]_0^{\pi/2} = (0 + 1) = 1.
      \]

   2. **Integrate with respect to \( r \):**
      We have:
      \[
      \int_0^2 r^2 e^{r \cos \theta} \, dr.
      \]
      To evaluate this, we would generally need integration by parts or a table of integrals, but given that it involves an exponential and polynomial term, it simplifies based on context or reference integral tables.

Thus, after evaluating, we can find the solution for the integral. Let me know if you'd like further details on the integration or the result. 

---

Would you like a deeper dive into any of these steps?

Here are five related questions for further exploration:
1. How do polar coordinates simplify the evaluation of double integrals in circular regions?
2. What other types of regions are best suited for polar coordinate transformations?
3. Can you apply the same technique for a circle of a different radius or a different quadrant?
4. What methods exist for evaluating more complex integrals involving exponential and trigonometric terms?
5. How does the use of symmetry help in evaluating integrals over specific regions?

**Tip:** When dealing with circular regions, switching to polar coordinates often makes the boundaries of the region easier to handle.

Evaluate the given integral by changing to polar coordinates: ∫∫_R y e^x dA, where R is the region in the first quadrant enclosed by the circle x^2 + y^2 = 4.

Learn how to evaluate the double integral ∫∫_R y e^x dA by converting to polar coordinates. This integral is defined over the region R, the first quadrant of the circle x^2 + y^2 = 4. A detailed step-by-step solution is provided for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation