Answer:
The Area of the Enclosed Region Is 64/3.
Step-by-step explanation:
As Given in Question
We have, Curve 4x+[tex]Y^{2}[/tex]=12
& X=Y
Solution.
4Y+[tex]Y^{2}[/tex]=12 (X=Y)
[tex]Y^{2}[/tex]+4Y-12=0
[tex]Y^{2}[/tex]+6Y-2Y-12=0
Y(Y+6)-2(Y+6)=0
(Y-2)*(Y+6)=0
Y=2 & -6 (X=Y)
Now at (2,2) & (-6,-6) both curves intersect each other.
The Area Of Enclosed Region is [tex]\int\limits^2_{-6} [(3-Y^{2}/4 )-Y] \, dy[/tex]
by Solving This Equation we get Area of Region = 64/3 .
this equation Solution & Curve Diagram please see In Attachment .
