What is the area enclosed by the curve `2x^(2)+y^(2)=1`?

Options

A. `2pi`
B. `pi`
C. `(pi)/2`
D. `(pi)/(sqrt(2))`

Answer

To find the area enclosed by the curve given by the equation \(2x^2 + y^2 = 1\), we can follow these steps:

### Step 1: Identify the type of curve
The equation \(2x^2 + y^2 = 1\) represents an ellipse.

### Step 2: Rewrite the equation in standard form
To rewrite the equation in the standard form of an ellipse, we can divide the entire equation by 1:
\[ \frac{x^2}{\frac{1}{2}} + \frac{y^2}{1} = 1 \]
This shows that the semi-major and semi-minor axes can be identified.

### Step 3: Identify the values of \(a\) and \(b\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can see:
- \(a^2 = \frac{1}{2} \Rightarrow a = \frac{1}{\sqrt{2}}\)
- \(b^2 = 1 \Rightarrow b = 1\)

### Step 4: Use the formula for the area of an ellipse
The area \(A\) of an ellipse is given by the formula:
\[ A = \pi \times a \times b \]
Substituting the values of \(a\) and \(b\):
\[ A = \pi \times \frac{1}{\sqrt{2}} \times 1 \]

### Step 5: Simplify the area
Thus, the area enclosed by the curve is:
\[ A = \frac{\pi}{\sqrt{2}} \text{ square units} \]

### Final Answer:
The area enclosed by the curve \(2x^2 + y^2 = 1\) is \(\frac{\pi}{\sqrt{2}}\) square units.

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