Answer:
[tex]Area = \infty[/tex]
Points of intersection: (1.225,-1.58) and (1.225,1.58)
Step-by-step explanation:
Given
[tex]x = \sqrt{4 - y^2}[/tex]
[tex]y^2 = 1 + x^2[/tex]
Required
The region area
Plot the graphs of [tex]x = \sqrt{4 - y^2}[/tex] and [tex]y^2 = 1 + x^2[/tex]
Make y the subject in both equations
[tex]x = \sqrt{4 - y^2}[/tex]
Square both sides
[tex]x^2 = 4 - y^2[/tex]
Rewrite
[tex]y^2 = 4 - x^2[/tex]
Take square roots
[tex]y = \sqrt{4 - x^2[/tex]
So, we have:
[tex]f(x) = \sqrt{4 - x^2}[/tex]
[tex]y^2 = 1 + x^2[/tex]
Take square roots
[tex]y = \sqrt{1 + x^2}[/tex]
So, we have:
[tex]g(x) = \sqrt{1 + x^2}[/tex]
The point of intersection is:
[tex]f(x) = g(x)[/tex]
[tex]\sqrt{4 - x^2} = \sqrt{1 + x^2}[/tex]
Square both sides
[tex]4 - x^2 = 1 + x^2[/tex]
Collect like terms
[tex]x^2 + x^2 = 4 - 1[/tex]
[tex]2x^2 = 3[/tex]
Divide by 2
[tex]x^2 = 1.5[/tex]
Take square roots
[tex]x = 1.225[/tex]
[tex]g(x) = \sqrt{1 + x^2}[/tex]
[tex]g(x) = \sqrt{1 + 1.5}[/tex]
[tex]g(x) = \sqrt{2.5}[/tex]
[tex]g(x) = \±1.581[/tex]
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:
[tex]Area = \infty[/tex]
