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=3 sqrt x,y=3 and 2y+1x=4
Since the region is roughly triangular, with a horizontal top but a vertex underneath, either way the region has to be broken into two parts. Either
∫[0,1] 3-(2-x/2) dx + ∫[1,4] 3-(3/2 √x) dx
or
∫[3/2,2] (4/9 y^2)-(4-2y) dy + ∫[2,3] 4/9 y^2 dy
To sketch the region enclosed by the given curves, we need to analyze the equations and their intersection points. The given curves are:
1) 2y = 3√x (equation 1)
2) y = 3 (equation 2)
3) 2y + x = 4 (equation 3)
To decide whether to integrate with respect to x or y, we need to see which variable the curves represent as their independent variable. Let's analyze each curve separately.
Curve 1:
Rewrite equation 1 to solve for y:
2y = 3√x
y = 3√(x/2)
From this equation, we can see that y depends on x. Therefore, we will integrate with respect to x.
Curve 2:
The equation y = 3 shows that y is constant at 3. So, y does not depend on x. Therefore, integrating with respect to y is not necessary for this curve.
Curve 3:
Rewrite equation 3 to solve for y:
2y = 4 - x
y = (4 - x)/2
y = 2 - (x/2)
In this equation, we can see that y depends on x. Therefore, we will integrate with respect to x.
Next, let's find the intersection points of these curves:
To find the intersection point between equations 1 and 2, we equate them and solve for x:
3√(x/2) = 3
√(x/2) = 1
x/2 = 1
x = 2
So, the intersection point is (2, 3).
To find the intersection point between equations 2 and 3, we equate them and solve for x:
2 - (x/2) = 3
x/2 = -1
x = -2
So, the intersection point is (-2, 3).
Now, let's sketch the region by plotting the curves and the intersection points on a graph:
First, sketch equation 1: y = 3√(x/2)
- Plot (2, 3), the intersection point with equation 2.
- Plot (0, 0) as a reference point.
- Sketch the curve smoothly passing through these points.
Second, sketch equation 2: y = 3
- Plot (2, 3) as a point.
- Draw a horizontal line passing through the point (2, 3).
Third, sketch equation 3: y = 2 - (x/2)
- Plot (-2, 3), the intersection point with equation 2.
- Plot (0, 2) as a reference point.
- Sketch the line smoothly passing through these points.
Now, we have outlined the enclosed region on the graph. It is a trapezoid with one side parallel to the x-axis and the other side following the shape of the curve in equation 1.
To find the area of this region, we will integrate with respect to x between the appropriate limits. The boundaries of the region are the intersection points (-2, 3) and (2, 3).
The area can be calculated by the integral:
A = ∫[from -2 to 2] (2 - (x/2)) - (3√(x/2)) dx
Evaluating this integral will give us the area of the region enclosed by the given curves.