Question

My fellow math brodas, help

Answer 1

A. Depending on which variable you choose to integrate with, you can capture the total bounded region with either -2 ≤ x ≤ (-1 + √5)/2 or 1 ≤ y ≤ (5 + √5)/2, where the upper endpoints correspond to the coordinates of the appropriate intersections:

y = x² + 1

⇒ x = (x² - 2)² - 2

⇒ x⁴ - 4x² - x + 2 = 0

⇒ (x - 2) (x + 1) (x² + x - 1) = 0

⇒ x = 2, x = -1, x = -1/2 ± √5/2

⇒ y = 5, y = 2, y = (5 ± √5)/2

On the other hand, we can compute the areas of A and B separately, then sum those integrals. Area A is easier to compute by integrating with respect to y over 2 ≤ y ≤ (5 + √5)/2, while area B is easier to find by integrating x over -1 ≤ x ≤ (-1 + √5)/2.

B. I'll stick to the split-region approach.

First, we find equations for the appropriates halves of either parabola:

• y = x² + 1 ⇒ x = ± √(y - 1)

and x = -√(y - 1) describes the left half of the blue parabola;

• x = (y - 3)² - 2 ⇒ y = 3 ± √(x + 2)

and y = 3 - √(x + 2) describe the bottom half of the red parabola.

Now we can set up the integrals.

Area of A:

`\displaystyle \int_2^((5+\sqrt5)/2) \left(\left(-√(y-1)\right) - \left((y-3)^2-2\right)\right) \, dy \\ ~~~~~~~~ = -\int_2^((5+\sqrt5)/2) \left((y-3)^2 - 2 + √(y-1)\right) \, dy`

Area of B:

`\displaystyle \int_(-1)^((-1+\sqrt5)/2) \left(\left(3-√(x+2)\right) - \left(x^2+1\right) \right) \, dx \\ ~~~~~~~~ = - \int_(-1)^((-1+\sqrt5)/2) \left(x^2 - 2 + √(x+2)\right) \, dx`

Alternatively, one can prove that the regions A and B are symmetric across the line y = x + 3, so we can simply pick one of these integrals and double it.

C. Computing the integrals, we find

area of A = 2/3

area of B = 2/3

and so the total area is 2/3 + 2/3 = 4/3.