Question

For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region.

x= √4-y^2 and y^2=1+x^2

Answer 1

`Area = \infty`

Points of intersection: (1.225,-1.58) and (1.225,1.58)

Explanation:

Given

`x = √(4 - y^2)`

`y^2 = 1 + x^2`

Required

The region area

Plot the graphs of
`x = √(4 - y^2)` and
`y^2 = 1 + x^2`

Make y the subject in both equations

`x = √(4 - y^2)`

Square both sides

`x^2 = 4 - y^2`

Rewrite

`y^2 = 4 - x^2`

Take square roots

`y = \sqrt{4 - x^2`

So, we have:

`f(x) = √(4 - x^2)`

`y^2 = 1 + x^2`

Take square roots

`y = √(1 + x^2)`

So, we have:

`g(x) = √(1 + x^2)`

The point of intersection is:

`f(x) = g(x)`

`√(4 - x^2) = √(1 + x^2)`

Square both sides

`4 - x^2 = 1 + x^2`

Collect like terms

`x^2 + x^2 = 4 - 1`

`2x^2 = 3`

Divide by 2

`x^2 = 1.5`

Take square roots

`x = 1.225`

`g(x) = √(1 + x^2)`

`g(x) = √(1 + 1.5)`

`g(x) = √(2.5)`

`g(x) = \±1.581`

So, the point of intersection is at: (1.225,-1.58) and (1.225,1.58)

See attachment

From the attached image, we can see that the curves do not enclose

Hence:

`Area = \infty`