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 90.
Did you make a sketch?
Did you notice that one parabola is the reflection of the other in the x-axis.
Lots of symmetry here
The 2 axes cut the region into 4 equal parts , each one would then be 22.5
Did you notice the x-intercept on the right side is (c,0) ?
okay then,
∫(c^2 - x^2) dx from 0 to c = 22.5
the integral is c^2 x - (1/3)x^3
so
c^3 - (1/3)c^3 = 22.5
(2/3) c^3 = 45/2
c^3 = 135/4 or 270/8 , (looking for perfect cubes, only got one)
c = (270/8)^(1/3)
= 3(cuberoot(10)/2 , (270 = 270x10 , 27 is a perfect cube)
I get appr 3.23
To find the value of c, we need to determine the points of intersection between the two parabolas.
First, let's set the two equations equal to each other:
x^2 - c^2 = c^2 - x^2
Simplifying this equation, we get:
2x^2 = 2c^2
Dividing both sides by 2, we have:
x^2 = c^2
Taking the square root of both sides, we get:
x = ±c
So, the two parabolas intersect at the points (c, c) and (-c, c).
To find the area of the region enclosed by the parabolas, we need to integrate the difference of the two equations over the interval [-c, c].
The area can be calculated using the definite integral:
Area = ∫[a, b] (f(x) - g(x)) dx
In this case, f(x) represents the upper parabola y = x^2 - c^2, and g(x) represents the lower parabola y = c^2 - x^2. The limits of integration, a and b, are -c and c because those are the points of intersection.
So, we have:
Area = ∫[-c, c] ((x^2 - c^2) - (c^2 - x^2)) dx
Simplifying this equation, we get:
Area = ∫[-c, c] (2x^2 - 2c^2) dx
Now, we can integrate the equation to find the area:
Area = [2/3 * x^3 - 2c^2x] from -c to c
Plugging in the limits of integration, we have:
Area = (2/3 * c^3 - 2c^3) - (-2/3 * c^3 + 2c^3)
Simplifying further, we get:
Area = (2/3 * c^3 - 2/3 * c^3) - (-2/3 * c^3 + 2/3 * c^3)
This simplifies to:
Area = (4/3 * c^3) - (4/3 * c^3)
The area is simply 0.
Therefore, there is no value of c that makes the enclosed area equal to 90.