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 and find the area, we need to understand the shape of each curve and their intersection points.

1. Given curve: 2y = 3√x
Rewrite this equation to isolate y:
y = (3/2)√x
This curve represents a square root function with a horizontal compression factor of (3/2). It is symmetric about the y-axis.
To find the x-intercept, plug in y = 0 and solve for x:
0 = (3/2)√x
0 = √x
x = 0
So, (0,0) is a point on the curve.

2. Given curve: y = 3
This curve is a horizontal line that intersects the y-axis at y = 3.
Since it is a line parallel to the x-axis, it intersects the x-axis at all possible x values.

3. Given curve: 2y + 3x = 6
Rewrite this equation to isolate y:
y = (6 - 3x)/2
This curve represents a linear function with a negative slope of -3/2.
To find the x-intercept, plug in y = 0 and solve for x:
0 = (6 - 3x)/2
0 = 6 - 3x
3x = 6
x = 2
So, (2,0) is a point on the curve.

Now, let's plot these curves on a coordinate plane:

- Draw a vertical line at x = 0 (y-axis).
- Plot the curve 2y = 3√x as a symmetric square root function curve opening to the right. Include the point (0,0) on the curve.
- Draw a horizontal line at y = 3.
- Plot the curve 2y + 3x = 6 as a linear function with a negative slope. Include the point (2,0) on the curve.

The region enclosed by the curves is a closed figure formed by these curves.

To find the area of the region, we need to determine the limits of integration. Since the curves are intersecting horizontally at y = 3, we will integrate with respect to y.

The lower limit of integration (y-value) is 0, as the curves intersect at this level on the y-axis. The upper limit of integration is 3, as the curves intersect at y = 3.

The formula to find the area of a region is given by:
A = ∫[lower limit, upper limit] f(y) dy

Since we integrate with respect to y, we need to express x in terms of y for each curve and find the appropriate integrands.

For the curve 2y = 3√x, square both sides to remove the square root:
(2y)^2 = (3√x)^2
4y^2 = 9x
Solving for x, we get:
x = (4y^2)/9

For the curve 2y + 3x = 6, rearrange the equation to solve for x:
3x = 6 - 2y
x = (6 - 2y)/3

Now, we can set up the integral to find the area of the region:
A = ∫[0,3] [(6 - 2y)/3 - (4y^2)/9] dy

Evaluating this integral will give us the area of the region enclosed by the given curves.