Find the area of the region enclosed by the given curves:
4x+y^2=9, x=2y
First, find where they intersect.
4x+y^2=9, x=2y
4(2y) + y^2 = 9
y^2 + 8y - 9 = 0
(y+9)(y-1) = 0
So, it might be easier to integrate in the y-direction:
∫[-9,1] (y^2-9)/4 - 2y dy
∫[-9,1] y^2/4 - 2y - 9/4 dy
y^3/12 - y^2 - 9/4 y [-9,1]
(-729/12 - 81 + 81/4) - (1/12 - 1 - 9/4)
355/3
Unfortunately, that answer is not right...
This is the last problem on my homework and I've been stuck on it for about an hour. Its very frustrating
How about 125/3 ?
If you had read my posting carefully (as I did not) you'd have noticed that it should have been
(9-y^2)/4
To find the area of the region enclosed by the given curves, we need to first determine the points of intersection between the two curves.
Let's start by solving the given system of equations:
1. 4x + y^2 = 9
2. x = 2y
Substituting equation 2 into equation 1:
4(2y) + y^2 = 9
8y + y^2 = 9
y^2 + 8y - 9 = 0
Now we can solve this quadratic equation for y by factoring or using the quadratic formula.
Factoring:
(y + 9)(y - 1) = 0
So we have two possible values for y: y = -9 and y = 1.
Now substitute these values back into equation 2 to find the corresponding x-values:
For y = -9:
x = 2(-9)
x = -18
For y = 1:
x = 2(1)
x = 2
Therefore, the two points of intersection are (-18, -9) and (2, 1).
To find the area of the region enclosed by the curves, we need to integrate the difference between the top curve and the bottom curve with respect to x over the x-range of the intersection points.
Since it is evident that the equation x = 2y represents the top curve and 4x + y^2 = 9 represents the bottom curve, we can set up the integral as follows:
∫[x1,x2] (top curve - bottom curve) dx
Where [x1, x2] represents the range of x-values between the points of intersection (-18, -9) and (2, 1).
Now, let's write the integral:
Area = ∫[-18, 2] ((2y) - (4x + y^2)) dx
To eliminate y from the equation, we can rewrite y in terms of x using equation 2:
y = x/2
Substituting this into the integral:
Area = ∫[-18, 2] ((2(x/2)) - (4x + (x/2)^2)) dx
Area = ∫[-18, 2] (x - (4x + (x^2/4))) dx
Area = ∫[-18, 2] (x - 4x - (x^2/4)) dx
Now, simplify the integrand:
Area = ∫[-18, 2] (-(3x + x^2/4)) dx
Finally, evaluate the definite integral from -18 to 2 to find the area of the region enclosed by the curves:
Area = ∫[-18, 2] (-(3x + x^2/4)) dx = [Insert the result of the integration]