Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a mathematician, she suggests building fences described by y=4x^2 and x^2+10. Farmer Jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife. What is the area of the enclosed region?
well, the two curves intersect at (±√(10/3),40/3), so the area is
2*integral[0,√(10/3)] (x^2+10)-(4x^2) dx = (40/3)√(10/3)
10
To find the area of the enclosed region, we need to find the points where the curves intersect. Setting the two equations equal to each other:
4x^2 = x^2 + 10
Simplifying:
3x^2 = 10
Dividing both sides by 3:
x^2 = 10/3
Taking the square root of both sides:
x = ± √(10/3)
Since we're looking for the area of the enclosed region, we only consider positive values for x. Therefore, x = √(10/3).
Substituting this value back into either equation, let's use y = 4x^2:
y = 4(√(10/3))^2
y = 4(10/3)
y = 40/3
So the coordinates of the point of intersection are (√(10/3), 40/3).
Next, let's find the x-coordinate of the other point where the curves intersect. We can do this by solving the equation:
x^2 + 10 = 4x^2
Rearranging the equation:
3x^2 - x^2 = 10
2x^2 = 10
Dividing both sides by 2:
x^2 = 5
Taking the square root of both sides:
x = ± √5
Again, we're only interested in positive values, so x = √5.
Substituting this value back into y = 4x^2:
y = 4(√5)^2
y = 4 x 5
y = 20
So the coordinates of the second point of intersection are (√5, 20).
Now, we can calculate the area of the enclosed region using the definite integral. The area between the curves is given by:
A = ∫(a to b) [f(x) - g(x)] dx
Where f(x) is the curve y = 4x^2, g(x) is the curve y = x^2 + 10, and a and b are the x-coordinates of the points of intersection.
Therefore, the area of the enclosed region is:
A = ∫(√(10/3) to √5) [(4x^2) - (x^2 + 10)] dx
Simplifying the integrand:
A = ∫(√(10/3) to √5) (3x^2 - 10) dx
Integrating:
A = [x^3 - 10x] from √(10/3) to √5
Evaluating the integral:
A = (√5^3 - 10√5) - ((√(10/3))^3 - 10(√(10/3)))
Simplifying:
A = (5√5 - 10√5) - ((10/3)√(10/3) - (10/3)√(10/3))
A = -5√5 + 10√5 + (10/3)√(10/3) - (10/3)√(10/3)
Simplifying further:
A = 5√5 + (10/3)√(10/3) - (10/3)√(10/3)
A = 5√5
Therefore, the area of the enclosed region is 5√5.
To find the area of the enclosed region, we need to calculate the integral of the function that represents the upper curve minus the lower curve.
In this case, the upper curve is given by y = 4x^2, and the lower curve is given by y = x^2 + 10.
To find the points of intersection between these two curves, we can set them equal to each other:
4x^2 = x^2 + 10
Subtracting x^2 from both sides, we get:
3x^2 = 10
Dividing by 3, we find:
x^2 = 10/3
Taking the square root of both sides, we get:
x = ±√(10/3)
Since we are interested in the positive values of x, we have:
x = √(10/3)
To calculate the area, we integrate the difference between the upper and lower curves with respect to x, over the interval [0, √(10/3)]:
Area = ∫[0, √(10/3)] (4x^2 - (x^2 + 10)) dx
Simplifying the integrand, we have:
Area = ∫[0, √(10/3)] (3x^2 - 10) dx
Integrating term by term, we get:
Area = [(x^3 - 10x)]|[0, √(10/3)]
Substituting the limits of integration, we have:
Area = [(√(10/3)^3 - 10√(10/3)) - (0 - 0)]
Simplifying further, we find:
Area = [(10√10/3) - 10√(10/3)]
Thus, the area of the enclosed region is (10√10/3) - 10√(10/3).