Sketch the region `{(x,y): 4x ^(2) + 9y ^(2) = 36}` and find its area, using integration.

Answer

To solve the problem of sketching the region defined by the equation \(4x^2 + 9y^2 = 36\) and finding its area using integration, we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ 4x^2 + 9y^2 = 36 \]
Divide every term by 36 to get it in standard form:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

### Step 2: Identify the Ellipse Parameters
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 4 \Rightarrow b = 2\)

Since \(a > b\), this is a horizontally oriented ellipse.

### Step 3: Sketch the Ellipse
The ellipse has its center at the origin \((0,0)\) with:
- Semi-major axis \(a = 3\) along the x-axis, extending to points \((3, 0)\) and \((-3, 0)\).
- Semi-minor axis \(b = 2\) along the y-axis, extending to points \((0, 2)\) and \((0, -2)\).

### Step 4: Set Up the Integral for Area Calculation
To find the area of the ellipse, we can calculate the area in the first quadrant and then multiply by 4. The area in the first quadrant is given by:
\[ \text{Area} = \int_0^3 y \, dx \]
First, we need to express \(y\) in terms of \(x\) from the ellipse equation:
\[ 9y^2 = 36 - 4x^2 \implies y^2 = 4 - \frac{4}{9}x^2 \implies y = \frac{2}{3}\sqrt{9 - x^2} \]

### Step 5: Write the Integral
Now, the area in the first quadrant becomes:
\[ \text{Area}_{\text{1st quadrant}} = \int_0^3 \frac{2}{3}\sqrt{9 - x^2} \, dx \]

### Step 6: Solve the Integral
Factor out the constant:
\[ \text{Area}_{\text{1st quadrant}} = \frac{2}{3} \int_0^3 \sqrt{9 - x^2} \, dx \]
The integral \(\int \sqrt{a^2 - x^2} \, dx\) has a standard result:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
For \(a = 3\), we evaluate from 0 to 3:
\[ \int_0^3 \sqrt{9 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{-1}\left(\frac{x}{3}\right) \right]_0^3 \]
Calculating at the limits:
- At \(x = 3\):
\[ \frac{3}{2}\sqrt{9 - 9} + \frac{9}{2} \sin^{-1}(1) = 0 + \frac{9}{2} \cdot \frac{\pi}{2} = \frac{9\pi}{4} \]
- At \(x = 0\):
\[ \frac{0}{2}\sqrt{9} + \frac{9}{2} \sin^{-1}(0) = 0 \]
Thus, the integral evaluates to:
\[ \int_0^3 \sqrt{9 - x^2} \, dx = \frac{9\pi}{4} \]

### Step 7: Calculate the Area of the Ellipse
Now, substituting back:
\[ \text{Area}_{\text{1st quadrant}} = \frac{2}{3} \cdot \frac{9\pi}{4} = \frac{3\pi}{2} \]
The total area of the ellipse is:
\[ \text{Total Area} = 4 \cdot \frac{3\pi}{2} = 6\pi \]

### Final Answer
The area of the ellipse is:
\[ \text{Area} = 6\pi \text{ square units} \]