Calculate the area of the bounded region between the curves y^2=x and 3y = -3y + 9 ?
3y = -3y + 9 ???
Fix that.
Then, find where the two graphs intersect. Since the parabola opens to the right, it's easier to integrate along dy. The area is just a bunch of thin strips of height dy and length the distance between the two curves. Express that and integrate from bottom to top.
3y = -3x + 9
the curves intersect at y = (-1-√13)/2 and (-1+√13)/2
area is thus
∫[(-1-√13)/2,(-1+√13)/2] (-y+3)-(y^2) dy
To calculate the area of the bounded region between the curves, we need to find the points of intersection between the curves and then integrate the difference in their y-values over the region.
First, let's find the points of intersection of the given curves:
1. y^2 = x
To find the intersection points, we need to equate the two equations:
y^2 = x --> x = y^2
2. 3y = -3y + 9
Simplifying this equation, we get:
3y + 3y = 9 --> 6y = 9 --> y = 9/6 --> y = 3/2
Now, we have the y-coordinate of the point of intersection between the curves. Let's find the x-coordinate:
x = y^2
x = (3/2)^2
x = 9/4
So, the point of intersection is (9/4, 3/2).
Now that we have the points of intersection, we can set up the integral to find the area. Since the given curves are y^2 = x and 3y = -3y + 9, the region will be bounded between y = 0 and y = 3/2.
The integral to calculate the area is given by:
Area = ∫[y=0 to y=3/2] (f(x) - g(x)) dy
Where f(x) is the upper curve and g(x) is the lower curve.
For this problem, the upper curve is y^2 = x and the lower curve is 3y = -3y + 9.
Let's solve the integral to find the area:
Area = ∫[y=0 to y=3/2] (y^2 - (9 - 3y)/3) dy
Simplifying the integrand, we get:
Area = ∫[y=0 to y=3/2] (y^2 - 3 + y) dy
Now, integrate the above expression with respect to y over the given bounds to find the area of the bounded region.