find the area of the region for
2y=4sqrt(x) , y=3, and 2y+3x=7
also help with
Find c>0 such that the area of the region enclosed by the parabolas
y=x^2-c^2 and y=c^2-x^2 is 430.
nice symmetry, should be easy to see that the x-intercepts are (c,0) and (-c,0) for both
so just double the area from 0 to c
area = 2∫(c^2 - x^2 - (x^2 - c^2)) dx from 0 to c
= 2∫(2c^2 - 2x^2) dx from 0 to c
= 2[2c^2 x - (2/3)x^3] from 0 to c
= 2(c^3 - (2/3)c^3]
= 2(1/3)c^3
= 2/3 c^3
but 2/3 c^3 = 430
c^3 = 645
c = ∛645 = appr 8.64
check my arithmetic
To find the area of the region enclosed by the given equations, we first need to determine the points of intersection between the curves.
Let's start with the first equation, 2y = 4√(x). To find the corresponding x-values, we need to isolate the variable x. Firstly, divide both sides of the equation by 2 to get y = 2√(x).
Now, let's consider the second equation, y = 3. This is a horizontal line passing through the y-coordinate 3.
Finally, let's look at the third equation, 2y + 3x = 7. To solve for y, subtract 3x from both sides and divide by 2, giving y = (7 - 3x) / 2. This equation represents a straight line.
Now, we have the equations y = 2√(x), y = 3, and y = (7 - 3x) / 2.
To find the points of intersection, we can equate these equations and solve for x and y.
1. Equating y = 2√(x) and y = 3, we get:
2√(x) = 3
Squaring both sides:
4x = 9
Solving for x:
x = 9/4
Substituting this value into the equation y = 3:
y = 3
Hence, the first point of intersection is (9/4, 3).
2. Equating y = 3 and y = (7 - 3x) / 2, we get:
3 = (7 - 3x) / 2
Multiplying both sides by 2:
6 = 7 - 3x
Rearranging:
3x = 1
Solving for x:
x = 1/3
Substituting this value into the equation y = 3:
y = 3
Hence, the second point of intersection is (1/3, 3).
Now that we have determined the points of intersection, we can integrate to find the area of the region enclosed by the curves.
The required area can be split into two regions, as shown:
1. The region between the curves y = 2√(x) and y = 3, from x = 0 to x = 9/4.
2. The region between the curves y = 3 and y = (7 - 3x) / 2, from x = 9/4 to x = 1/3.
The area of each region can be calculated separately using definite integration.
1. The area of the region between y = 2√(x) and y = 3, from x = 0 to x = 9/4, can be found using the integral:
∫(3 - 2√(x)) dx, with limits of integration from 0 to 9/4.
2. The area of the region between y = 3 and y = (7 - 3x) / 2, from x = 9/4 to x = 1/3, can be found using the integral:
∫((7 - 3x)/2 - 3) dx, with limits of integration from 9/4 to 1/3.
Evaluating these integrals will give you the areas of the respective regions, and their sum will be the total area of the given region.
3 graphs:
y = 2√x
y = 3
y = (-3/2)x + 7
see:
http://www.wolframalpha.com/input/?i=plot++y+%3D+2√x+%2C+y+%3D+3+%2C+y+%3D+%28-3%2F2%29x+%2B+7+%2C+2%3Cx%3C3
intersection of y = (-3/2)x+7 and y = 2√x
is aprr A(2.54, 3.188)
intersection of y = 2√x and y = 3
is B(9/4, 3)
intersection of y = (-3/2)x + 7 and y = 3 is
C(8/3, 3)
So we want the "trianglular" shaped region ABC
From A, draw a perpendicualar to meet y = 3 at P
So APC is a right-angled triangle
area is easy to find
for the other
area = ∫(2√x - 3) dx from x = 9/4 to 2/54
looks routine, messy arithmetic.