Question

1. x = y² and x = 3 - 2y²

Explanation:

Assuming the problem is to find the area between the curves \( x = y^2 \) and \( x = 3 - 2y^2 \), here's the solution:

Step 1: Find Intersection Points

Set \( y^2 = 3 - 2y^2 \) to find where the curves intersect.
\( y^2 + 2y^2 = 3 \)
\( 3y^2 = 3 \)
\( y^2 = 1 \)
\( y = \pm 1 \)

Step 2: Determine the Right and Left Curves

For a given \( y \), compare \( x = y^2 \) and \( x = 3 - 2y^2 \).
Since \( 3 - 2y^2 \geq y^2 \) for \( y \in [-1, 1] \) (test \( y = 0 \): \( 3 \geq 0 \)), the right curve is \( x = 3 - 2y^2 \) and the left curve is \( x = y^2 \).

Step 3: Set Up the Integral for Area

The area \( A \) between two curves \( x = f(y) \) and \( x = g(y) \) from \( y = a \) to \( y = b \) is \( A = \int_{a}^{b} [f(y) - g(y)] dy \).
Here, \( f(y) = 3 - 2y^2 \), \( g(y) = y^2 \), \( a = -1 \), \( b = 1 \).
\( A = \int_{-1}^{1} [(3 - 2y^2) - y^2] dy \)
Simplify the integrand: \( 3 - 3y^2 \)

Step 4: Integrate and Evaluate

Since the integrand is even (symmetric about \( y = 0 \)), we can integrate from \( 0 \) to \( 1 \) and double it.
\( A = 2 \int_{0}^{1} (3 - 3y^2) dy \)
Integrate term by term:
\( \int (3 - 3y^2) dy = 3y - y^3 + C \)
Evaluate from \( 0 \) to \( 1 \):
\( [3(1) - (1)^3] - [3(0) - (0)^3] = 3 - 1 = 2 \)
Multiply by 2: \( 2 \times 2 = 4 \)

Answer:

The area between the curves is \(\boldsymbol{4}\).