Find the areas of the regions bounded by the lines and curves by expressing x as a function of y and integrating with respect to y.
x = (y-1)² - 1, x = (y-1)² + 1 from y=0 to y=2.
I graphed the two functions and the do not intersect? Does it matter? Or do I still find the area in between?
Thank you!
You're right, unless you have made a typo, the two curves will never intersect.
As you said, it does not matter, the lines y=0 and y=2 will intersect both lines to make a curved rectangle whose area you'd have to calculate. The result should be a nice integer.
Even though they don't intersect, there is the area bounded between y=0 and y=2
Did you notice that the two parabolas are congruent and the second is merely translated 2 units to the right?
so the horizontal distance between corresponding points is always 2
that is x2-x1 = (y-1)^2 + 1 - ((x-1)^2 - 1) = 2
Area = ∫x dy from 0 to 2
= ∫ 2dy from 0 to 2
= [2y] from 0 to 2
= 4-0 = 4
check my thinking, seems too easy.
To find the areas of the regions bounded by the given lines and curves, first, we need to determine the points at which they intersect. In this case, you correctly observed that the two functions do not intersect.
However, even if the functions do not intersect, it is still possible to find the area between them by taking the absolute difference between the functions at each y-value and integrating over the interval where the lines and curves are defined.
In this case, the two functions are x = (y-1)² - 1 and x = (y-1)² + 1. The interval of integration is from y=0 to y=2.
To proceed with finding the areas, we need to express x as a function of y for each equation. For the first equation, x = (y-1)² - 1, we can solve for y:
(x+1) = (y-1)²
√(x+1) = y - 1
y = √(x+1) + 1
Similarly, for the second equation, x = (y-1)² + 1, we can solve for y:
(x-1) = (y-1)²
√(x-1) = y - 1
y = √(x-1) + 1
Now, the area between the two functions can be found by integrating the absolute difference between these two expressions for y with respect to y:
Area = ∫[0,2] |√(x+1) + 1 - √(x-1) - 1| dy
This integral can be evaluated numerically or simplified further using algebraic techniques.