Sketch the given region of integration R and evaluate the integral over R using polar coordinates.

ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper R StartFraction 1 Over StartRoot 36 minus x squared minus y squared EndRoot EndFraction dA∫∫R136−x2−y2 dA​;

Requals=StartSet left parenthesis x comma y right parenthesis : x squared plus y squared less than or equals 1 comma x greater than or equals 0 comma y greater than or equals 0 EndSet

Let's walk through the solution for evaluating the integral over the given region \( R \) by using **polar coordinates**.

---

### **1. Understand the region \( R \):**

The region of integration \( R \) is defined as:  
\[
R = \left\{ (x, y) : x^2 + y^2 \leq 1, \quad x \geq 0, \quad y \geq 0 \right\}
\]

This describes the **quarter-circle** in the **first quadrant** with radius \( 1 \). That is, it includes points \( (x, y) \) such that the distance from the origin is at most 1, but only in the first quadrant.

---

### **2. Set up the integral in polar coordinates:**

We convert the integral to polar coordinates, using:
\[
x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta
\]
For a **quarter-circle** in the first quadrant:
- **\( r \)** ranges from \( 0 \) to \( 1 \).
- **\( \theta \)** ranges from \( 0 \) to \( \frac{\pi}{2} \).

Now, we rewrite the integrand.  
The integrand is:
\[
\frac{1}{\sqrt{36 - x^2 - y^2}}
\]
In polar coordinates:
\[
x^2 + y^2 = r^2 \quad \Rightarrow \quad 36 - x^2 - y^2 = 36 - r^2
\]
Thus, the integrand becomes:
\[
\frac{1}{\sqrt{36 - r^2}}
\]

---

### **3. Rewrite the integral:**

The original integral over \( R \) becomes:
\[
\int_0^{\pi/2} \int_0^1 \frac{1}{\sqrt{36 - r^2}} \, r \, dr \, d\theta
\]

---

### **4. Evaluate the integral:**

#### Step 1: Integrate with respect to \( r \).

We need to compute:
\[
\int_0^1 \frac{r}{\sqrt{36 - r^2}} \, dr
\]
Use the **substitution**:  
\( u = 36 - r^2 \quad \Rightarrow \quad du = -2r \, dr \).  
Thus:
\[
\int_0^1 \frac{r}{\sqrt{36 - r^2}} \, dr = -\frac{1}{2} \int_{36}^{35} \frac{-du}{\sqrt{u}} = \frac{1}{2} \int_{35}^{36} u^{-1/2} \, du
\]
Evaluating the integral:
\[
\frac{1}{2} \left[ 2u^{1/2} \right]_{35}^{36} = \left[ \sqrt{u} \right]_{35}^{36} = \sqrt{36} - \sqrt{35} = 6 - \sqrt{35}
\]

#### Step 2: Integrate with respect to \( \theta \).

Now we integrate with respect to \( \theta \):
\[
\int_0^{\pi/2} (6 - \sqrt{35}) \, d\theta = (6 - \sqrt{35}) \cdot \left[ \theta \right]_0^{\pi/2} = (6 - \sqrt{35}) \cdot \frac{\pi}{2}
\]

---

### **5. Final Answer:**

The value of the integral is:
\[
\frac{\pi}{2} (6 - \sqrt{35})
\]

---

### **Summary:**

The integral evaluates to:
\[
\frac{\pi}{2} (6 - \sqrt{35})
\]

---

Would you like me to provide more detailed steps or further clarifications?

---

### **Related Questions:**

1. How do you convert an integral from Cartesian to polar coordinates?
2. What is the geometric meaning of the limits in polar coordinates?
3. How does the Jacobian factor \( r \) appear in polar coordinates?
4. How do you integrate functions involving \( \sqrt{a^2 - r^2} \)?
5. What happens if the region of integration covers a full circle instead of a quadrant?

---

### **Tip:**  
When switching to polar coordinates, remember that the area element changes from \( dx \, dy \) to \( r \, dr \, d\theta \), which accounts for the stretching near the origin.

Sketch the given region of integration R and evaluate the integral over R using polar coordinates.
∫∫_R (1 / √(36 - x^2 - y^2)) dA;
R = {(x, y) : x^2 + y^2 ≤ 1, x ≥ 0, y ≥ 0}

Discover how to convert a double integral from Cartesian to polar coordinates and evaluate it over a quarter-circle region in the first quadrant. This step-by-step solution covers key calculus concepts, such as double integrals, polar transformations, and the use of Jacobians. Perfect for students tackling college-level multivariable calculus problems.

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