## Dido525 2 years ago Find the area of the plane region that is enclosed by the curves: $y=\left| x^2-4 \right|$ and$y=(\frac{ x^2 }{ 2 }) +4$

1. Dido525

I just need to know what to do with the absolute value sign.

2. Dido525

@SmoothMath @Zarkon @UnkleRhaukus

3. Dido525

@Hero

4. Hero

You graph it. Remember y = |x^2 -4| = x^2 - 4 or 4 - x^2

5. Dido525

So it's really 3 curves I am dealing with right?

6. Hero

You have to pay attention to the OR part. You will have two separate graphs. However, my theory is that the areas of both will be the same.

7. Dido525

Okay now I am confused.

8. Dido525

I get that i consider the or part but then how do I find the area?

9. Hero

On one graph, you will find the area enclosing y = x^2 - 4 and y = x^2/2 + 4 On the other graph you will find the are enclosing y = 4 - x^2 and y = x^2/2 + 4

10. Dido525

Ohh okay! So I find the area between the curves for both cases and add them up right?

11. Hero

In theory

12. Dido525

Hmm let me do it...

13. Hero

Don't forget to find the enclosing points first.

14. Dido525

Yeah that's easy.

15. Dido525

yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.

16. UnkleRhaukus

|dw:1355700012700:dw|

17. UnkleRhaukus

|dw:1355700073022:dw|

18. Dido525

Yep I got that... but you have to do this with a calculator don't you?

19. UnkleRhaukus

i would break the graph in half first , because it is symmetric about, the y axis then i would calculate the area when x is from 0 to 2, and add this to the area from x=2 to x=4

20. Dido525

I JUST noticed the symmetry :P .

21. UnkleRhaukus

|dw:1355700278738:dw|

22. Dido525

So I would go: |dw:1355700438412:dw|

23. Dido525

Right? To deal with:|dw:1355700518056:dw|

24. UnkleRhaukus

that looks right to me, (you draw your fours funny) but yes thats is the right integration

25. Dido525

My teachers said my 4's looked like 9 >.< .

26. UnkleRhaukus

looks more like a five to me

27. Dido525

|dw:1355700680945:dw|

28. UnkleRhaukus

so thats just where the first curve y=|x^2-4| is really y= 4-x^2

29. UnkleRhaukus

(like Hero was trying to saying )

30. Dido525

Ohh! So it would be: |dw:1355700839817:dw|

31. Dido525

Yeah I know hero said that :) .

32. UnkleRhaukus

yeah thats right ,

33. Dido525

Thanks a lot both of you :) @UnkleRhaukus @Hero .

34. Dido525

Could you give hero a medal too :) .

35. UnkleRhaukus

so take those integrals, add them together, then dont forget to Double your answer because we were only looking at the RHS of the graph

36. Dido525

Yep I know :) .

37. Hero

No @UnkleRhaukus. I wasn't trying to say it. I did say it.