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=3sqrtx , y=3 , 2y+2x=5
You will have to sketch the three curves yourself. The curves
y = (3/2)sqrtx and y = (5/2) -x
intersect at (1,3/2)
The enclosed area starts at that point and expands upward to the line y = 3, bounded by the two other curves along the way.
Your best bet is to integrate
(x2 - x1) =
(4/9)y^2 - [(5/2)-y]dy from y= 3/2 to 3
To sketch the region enclosed by the given curves, we need to find the points of intersection and the boundaries of the region.
First, let's find the points of intersection between the curves:
1. Curve 1: 2y = 3√x
To simplify, square both sides of the equation:
(2y)^2 = (3√x)^2
4y^2 = 9x
2. Curve 2: y = 3
This is a horizontal line at y = 3.
3. Curve 3: 2y + 2x = 5
Simplify the equation:
2y = 5 - 2x
y = (5 - 2x) / 2
Now, let's find the points of intersection:
1. Curve 1 and Curve 2:
Set the values of y equal to find the x-coordinate:
4(3^2) = 9x
36 = 9x
x = 4
Substituting x back into Curve 1, we get:
4y^2 = 9x
4y^2 = 9(4)
y^2 = 9
y = ±3
So, the point of intersection is (4, 3) and (4, -3).
2. Curve 2 and Curve 3:
Set the y-values equal to each other:
3 = (5 - 2x) / 2
Solve for x:
6 = 5 - 2x
2x = -1
x = -1/2
Substituting x back into Curve 3, we get:
y = (5 - 2x) / 2
y = (5 - 2(-1/2)) / 2
y = (5 + 1) / 2
y = 3
So, the point of intersection is (-1/2, 3).
Now that we have the points of intersection, we can draw the region enclosed by the curves:
- Curve 1 is a parabolic shape opening to the right.
- Curve 2 is a horizontal line at y = 3.
- Curve 3 is a linear function.
The region enclosed is bounded by Curve 1, Curve 2, and Curve 3.
To determine whether to integrate with respect to x or y, we need to examine the y-values of the curves.
- Curve 2 has a constant y-value of 3, indicating that it is a horizontal boundary.
- Curve 1 and Curve 3 both have a variable y-value. Curve 1 represents the lower boundary, and Curve 3 represents the upper boundary.
Based on the given information, we integrate with respect to y.
To find the area of the region, we will find the definite integral of the upper and lower boundaries:
Integral [Curve 3, Curve 1] (upper boundary - lower boundary) dy
We can integrate in terms of y since we determined we are integrating with respect to y.
Evaluate the integral and find the area of the region.