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=4*sqrt(x) , y=5 and 2y+4x=8

please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't arrive at the right area between the curves. Thanks in advance!

Thanks I figured it out my answer was 9, and it was correct

first of all reduce the equation to simplest form:

y = 2√x
y = 5
y + 2x = 4

Your sketch should look like a "triangle" with one side slightly curved
see:
http://www.wolframalpha.com/input/?i=plot+y%3D2*sqrt%28x%29+%2C+y%3D5+%2C+y%2B2x%3D4

we need all the intersection points,
I assume you know how to do that, ....
but I will do the hardest one:
intersect y = 2√x and y + 2x = 4
then 2√x = (4-2x)
or
√x = 2-x
by inspection, x = 1 , then in the originals , y = 2
point of intersection is (1,2)
Hoping that you get (-1/2, 5) and ( 25/4,5) for the other points

so area
= ∫(5 - (4-2x)) dx from -1/2 to 1 + ∫(5 - 2x^(1/2) dx from 1 to 25/4
= (x + x^2)| from -1/2 to 1 + (5x - 4x^(-1/2) )| from 1 to 25/4
= 1 + 1 - (-1/2 + 1/4) + (125/4 - 8/5 - (5-4) )
= 2 + 1/4 + 125/4 - 8/5 - 1
= 309/10

you better check my arithmetic on that one, should have written it out on paper first.

To sketch the region enclosed by the given curves and decide whether to integrate with respect to x or y, let's start by plotting the curves on a graph.

Given curves:
1) 2y = 4*√(x)
2) y = 5
3) 2y + 4x = 8

We can sketch the graph by representing y on the vertical axis and x on the horizontal axis.

1) Curve 1: 2y = 4*√(x)
We can rewrite this equation as y = 2*√(x), which represents a square root function. When x = 0, y = 0, and if we increase x, y will also increase. Plotting a few points (0,0), (1,2), (4,4), we can draw a curve that starts at the origin and curves upward.

2) Curve 2: y = 5
This is a horizontal line parallel to the x-axis, passing through y = 5.

3) Curve 3: 2y + 4x = 8
Rewriting this equation to y = -2x + 4 represents a straight line with a slope of -2 and y-intercept of 4. Plotting a few points, such as (0,4), (2,0), and (-2,8), we draw a line sloping downward.

We need to find the points of intersection between the curves to determine the boundaries of the region.

Setting curve 1 equal to curve 2, we have:
2*√(x) = 5

Squaring both sides, we get:
4x = 25

Simplifying, we find x = 25/4. Plugging this into curve 1, we have:
y = 2*√(25/4) = 5

Therefore, the region is bounded by the curves y = 5, y = 2*√(x), and y = -2x + 4, with x ranging from 0 to 25/4.

To find the area of the region, we need to integrate with respect to y.

The bounds for y are from y = 0 (the x-axis) to y = 5.

The expression for the outer curve is given by:
y = 5

The expression for the inner curve is given by:
y = 2*√(x)

Thus, the integral expression to find the area A is:
A = ∫[0 to 5] [2*√(x) - 5] dy

Simplifying, we have:
A = ∫[0 to 5] [2*√(x) - 5] dy

To evaluate this integral, we need to express the bounds in terms of x instead of y. Since we have the equation y = 2*√(x), we can rewrite it as x = (1/4)*y^2.

Substituting the bounds for y, we have:
A = ∫[0 to 5] [2*√((1/4)*y^2) - 5] dy
= ∫[0 to 5] [2*(1/2)*y - 5] dy
= ∫[0 to 5] [y - 5] dy

Evaluating this integral, we have:
A = [y^2/2 - 5y] from 0 to 5
= [(5)^2/2 - 5(5)] - [(0)^2/2 - 5(0)]
= [25/2 - 25] - [0 - 0]
= 25/2 - 25
= -25/2

The area of the region enclosed by the given curves is -25/2 square units.

Note: If you only consider the positive y-values for the area, the result will be 25/2 square units.

To sketch the region enclosed by the curves, we first need to find the points of intersection.

1. Start by setting the equations equal to each other:
2y = 4√(x) ----(1)
2y + 4x = 8 ----(2)

2. Solve equation (2) for y:
2y = 8 - 4x
y = 4 - 2x ----(3)

3. Now, equate equation (1) and equation (3) to find the points of intersection:
4 - 2x = 4√(x)

4. Square both sides of the equation to eliminate the square root:
(4 - 2x)^2 = (4√(x))^2
16 - 16x + 4x^2 = 16x (after expanding)

5. Simplify the equation:
4x^2 - 32x + 16 = 0

6. Divide the equation by 4 to simplify it further:
x^2 - 8x + 4 = 0

7. Solve for x using the quadratic formula:
x = [8 ± √(8^2 - 4(1)(4))] / (2(1))

Simplifying this equation further, we get:
x = [8 ± √(64 - 16)] / 2
x = [8 ± √48] / 2
x = 4 ± 2√3

Therefore, we have two values of x: x = 4 + 2√3 and x = 4 - 2√3.

8. Substitute these x-values into equation (3) to find the corresponding y-values:
For x = 4 + 2√3:
y = 4 - 2(4 + 2√3)
y = 4 - 8 - 4√3
y = -4 - 4√3

For x = 4 - 2√3:
y = 4 - 2(4 - 2√3)
y = 4 - 8 + 4√3
y = -4 + 4√3

Hence, the points of intersection are:
(4 + 2√3, -4 - 4√3) and (4 - 2√3, -4 + 4√3)

Next, let's determine whether to integrate with respect to x or y.

Looking at the given curves, we can see that the curve 2y = 4√(x) is equivalent to y = 2√(x), a parabolic function. Meanwhile, the curve 2y + 4x = 8 is a linear equation.

Since the region is enclosed between these curves, the graph will look like a "petal" shape. The left and right parts of the graph are both bounded by the parabola, while the middle part is bounded by the linear equation.

To find the area of the region, we need to divide it into two parts and integrate separately. We can choose to integrate either with respect to x or y, depending on which variable is easier to handle.

In this case, it is simpler to integrate with respect to y.

We can see that for y = -4 - 4√3 ≤ y ≤ 5, the region is bounded by the linear equation. And for y = -4 + 4√3 ≤ y ≤ 5, the region is bounded by the parabolic curve.

Let's start by finding the area between the linear equation and the x-axis.

1. Integrate the linear equation (2y + 4x = 8) with respect to y:
Solve the linear equation for x:
4x = 8 - 2y
x = 2 - 0.5y

2. Integrate x with respect to y within the given y-limits:
A1 = ∫[(-4 - 4√3), 5] (2 - 0.5y) dy

Now, let's find the area between the parabolic curve and the x-axis.

1. Integrate the parabolic curve (2y = 4√(x)) with respect to y:
Solve the parabolic curve for x:
x = (y/2)^2

2. Integrate x with respect to y within the given y-limits:
A2 = ∫[( -4 + 4√3), (-4 - 4√3)] (y/2)^2 dy

Finally, calculate the total area by adding A1 and A2 together.

Area = A1 + A2

You can use integration techniques to find the definite integrals, or you can use software or calculators that can perform definite integrals to obtain the numerical value of the area.