
#1
Jun1009, 05:38 PM

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Find c>0 such that the area of the region enclosed by the parabolas y= x^2c^2 and
y = c^2x^2 is 110 what is value of c i got 3.49 and it is wrong i dont understand how to do this problem any help would be greatly appreciated 



#3
Jun1009, 05:52 PM

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It's hard to tell what you did wrong if you don't show us how you got 3.49.




#4
Jun1009, 06:01 PM

P: 20

finding area of shaded regionx = + c , c a = 4 integral 0 to c (c^2x^2)dx 110 = 4 times [ c^2*x(1/3)x^3] limit 0 to c 110 = (8/3) c^3 c = 3.45 



#5
Jun1009, 06:01 PM

P: 20

x = + c , c a = 4 integral 0 to c (c^2x^2)dx 110 = 4 times [ c^2*x(1/3)x^3] limit 0 to c 110 = (8/3) c^3 c = 3.45 



#6
Jun1009, 06:21 PM

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I get 3.45... as well. But that's not the same as 3.49. Maybe they want you to express it exactly using a cube root?




#7
Jun1009, 07:18 PM

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#9
Jun1009, 09:05 PM

P: 20

no i tried all that and it still doesnt work is the math correct the formula i mean i dont even know if i am even doing the problem right




#10
Jun1009, 09:25 PM

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It looks fine to me.




#11
Jun1009, 10:38 PM

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#12
Jun1109, 03:47 PM

P: 20

this is what he did y = x^2 c^2 y = (0)^2C^2 y = c^2X^2 y = c^2(0) y = c^2 c^2X^2=x^2C^2 2c^2 = 2x^2 +c = X area integral a to b (y top  y bottom)dx 110 = integral c to c (c^2x^2)(x^2c^2) 



#13
Jun1109, 03:57 PM

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I think he is figuring out which curve is on top of the region (y=c^2x^2) and which is on the bottom (y=x^2c^2) and then the x boundaries +c and c. He then subtracts the lower curve y from the upper curve y and integrates over the whole region at once (rather than doing the first quadrant and multiplying by 4, like you did). But if you do his final integral, you get (8/3)*c^3. Just as you did.




#14
Jun1109, 04:10 PM

P: 20




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