Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=3sqrt(x) , y=3 and 2y+3x=6.

To sketch the region enclosed by the given curves, we first need to determine the intersection points.

1. The first curve is given by 2y = 3√(x).
Solving for y, we have y = (3/2)√(x).

2. The second curve is given by y = 3.

3. The third curve is given by 2y + 3x = 6.
Solving for y, we have y = 3 - (3/2)x.

Now, let's find the intersection points by setting the equations for y equal to each other:

(3/2)√(x) = 3 - (3/2)x

To solve this equation, we can square both sides to eliminate the square root:

(9/4)x = 9 - 9x + (9/4)x^2

Simplifying, we have:

(9/4)x^2 - (27/4)x + 9 = 0

Now, we can solve this quadratic equation. Taking out the common factor of (9/4), we have:

(9/4)(x^2 - 3x + 4) = 0

Since we can't solve this quadratic equation directly, let's use the quadratic formula:

x = (-(-3) ± √((-3)^2 - 4(1)(4)))/(2(1))

Simplifying, we have:

x = (3 ± √(9 - 16))/2
x = (3 ± √(-7))/2

Since we have a negative value under the square root, it means there are no real solutions for x.

Therefore, there are no intersection points, and hence, there is no enclosed region between the curves.

As a result, the area of the region is zero.