Dido525
Find the area of the plane region that is enclosed by the curves:
\[y=\left| x^2-4 \right|\] and\[y=(\frac{ x^2 }{ 2 }) +4\]
Delete
Share
This Question is Closed
Dido525
Best Response
You've already chosen the best response.
0
I just need to know what to do with the absolute value sign.
Dido525
Best Response
You've already chosen the best response.
0
@SmoothMath @Zarkon @UnkleRhaukus
Dido525
Best Response
You've already chosen the best response.
0
@Hero
Hero
Best Response
You've already chosen the best response.
1
You graph it. Remember
y = |x^2 -4| = x^2 - 4 or 4 - x^2
Dido525
Best Response
You've already chosen the best response.
0
So it's really 3 curves I am dealing with right?
Hero
Best Response
You've already chosen the best response.
1
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.
Dido525
Best Response
You've already chosen the best response.
0
Okay now I am confused.
Dido525
Best Response
You've already chosen the best response.
0
I get that i consider the or part but then how do I find the area?
Hero
Best Response
You've already chosen the best response.
1
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
Dido525
Best Response
You've already chosen the best response.
0
Ohh okay! So I find the area between the curves for both cases and add them up right?
Hero
Best Response
You've already chosen the best response.
1
In theory
Dido525
Best Response
You've already chosen the best response.
0
Hmm let me do it...
Hero
Best Response
You've already chosen the best response.
1
Don't forget to find the enclosing points first.
Dido525
Best Response
You've already chosen the best response.
0
Yeah that's easy.
Dido525
Best Response
You've already chosen the best response.
0
yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.
UnkleRhaukus
Best Response
You've already chosen the best response.
1
|dw:1355700012700:dw|
UnkleRhaukus
Best Response
You've already chosen the best response.
1
|dw:1355700073022:dw|
Dido525
Best Response
You've already chosen the best response.
0
Yep I got that... but you have to do this with a calculator don't you?
UnkleRhaukus
Best Response
You've already chosen the best response.
1
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
Dido525
Best Response
You've already chosen the best response.
0
I JUST noticed the symmetry :P .
UnkleRhaukus
Best Response
You've already chosen the best response.
1
|dw:1355700278738:dw|
Dido525
Best Response
You've already chosen the best response.
0
So I would go:
|dw:1355700438412:dw|
Dido525
Best Response
You've already chosen the best response.
0
Right? To deal with:|dw:1355700518056:dw|
UnkleRhaukus
Best Response
You've already chosen the best response.
1
that looks right to me, (you draw your fours funny) but yes thats is the right integration
Dido525
Best Response
You've already chosen the best response.
0
My teachers said my 4's looked like 9 >.< .
UnkleRhaukus
Best Response
You've already chosen the best response.
1
looks more like a five to me
Dido525
Best Response
You've already chosen the best response.
0
|dw:1355700680945:dw|
UnkleRhaukus
Best Response
You've already chosen the best response.
1
so thats just where the first curve
y=|x^2-4| is really y= 4-x^2
UnkleRhaukus
Best Response
You've already chosen the best response.
1
(like Hero was trying to saying )
Dido525
Best Response
You've already chosen the best response.
0
Ohh! So it would be:
|dw:1355700839817:dw|
Dido525
Best Response
You've already chosen the best response.
0
Yeah I know hero said that :) .
UnkleRhaukus
Best Response
You've already chosen the best response.
1
yeah thats right ,
Dido525
Best Response
You've already chosen the best response.
0
Thanks a lot both of you :) @UnkleRhaukus @Hero .
Dido525
Best Response
You've already chosen the best response.
0
Could you give hero a medal too :) .
UnkleRhaukus
Best Response
You've already chosen the best response.
1
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
Dido525
Best Response
You've already chosen the best response.
0
Yep I know :) .
Hero
Best Response
You've already chosen the best response.
1
No @UnkleRhaukus. I wasn't trying to say it. I did say it.