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

1. Dido525 Group Title

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

2. Dido525 Group Title

@SmoothMath @Zarkon @UnkleRhaukus

3. Dido525 Group Title

@Hero

4. Hero Group Title

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

5. Dido525 Group Title

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

6. Hero Group Title

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 Group Title

Okay now I am confused.

8. Dido525 Group Title

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

9. Hero Group Title

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 Group Title

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

11. Hero Group Title

In theory

12. Dido525 Group Title

Hmm let me do it...

13. Hero Group Title

Don't forget to find the enclosing points first.

14. Dido525 Group Title

Yeah that's easy.

15. Dido525 Group Title

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

16. UnkleRhaukus Group Title

|dw:1355700012700:dw|

17. UnkleRhaukus Group Title

|dw:1355700073022:dw|

18. Dido525 Group Title

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

19. UnkleRhaukus Group Title

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 Group Title

I JUST noticed the symmetry :P .

21. UnkleRhaukus Group Title

|dw:1355700278738:dw|

22. Dido525 Group Title

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

23. Dido525 Group Title

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

24. UnkleRhaukus Group Title

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

25. Dido525 Group Title

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

26. UnkleRhaukus Group Title

looks more like a five to me

27. Dido525 Group Title

|dw:1355700680945:dw|

28. UnkleRhaukus Group Title

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

29. UnkleRhaukus Group Title

(like Hero was trying to saying )

30. Dido525 Group Title

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

31. Dido525 Group Title

Yeah I know hero said that :) .

32. UnkleRhaukus Group Title

yeah thats right ,

33. Dido525 Group Title

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

34. Dido525 Group Title

Could you give hero a medal too :) .

35. UnkleRhaukus Group Title

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 Group Title

Yep I know :) .

37. Hero Group Title

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