What is the value of c for a shaded area of 110 between two parabolas?

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Find c>0 such that the area of the region enclosed by the parabolas y= x^2-c^2 and
y = c^2-x^2 is 110

what is value of c

i got 3.49 and it is wrong i don't understand how to do this problem any help would be greatly appreciated
 
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intelli said:
Find c>0 such that the area of the region enclosed by the parabolas y= x^2-c^2 and
y = c^2-x^2 is 110

what is value of c

i got 3.49 and it is wrong i don't understand how to do this problem any help would be greatly appreciated

How did you get 3.49? Show us your work please.
 
It's hard to tell what you did wrong if you don't show us how you got 3.49.
 
berkeman said:
How did you get 3.49? Show us your work please.

ok

x = + c , -c

a = 4 integral 0 to c (c^2-x^2)dx

110 = 4 times [ c^2*x-(1/3)x^3] limit 0 to c

110 = (8/3) c^3

c = 3.45
 
berkeman said:
How did you get 3.49? Show us your work please.

ok

x = + c , -c

a = 4 integral 0 to c (c^2-x^2)dx

110 = 4 times [ c^2*x-(1/3)x^3] limit 0 to c

110 = (8/3) c^3

c = 3.45
 
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?
 
Dick said:
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?

no i tried that it doesn't work
 
intelli said:
no i tried that it doesn't work

I get 3.4552116... That's not quite 3.45 to 3 sig figs.
 
no i tried all that and it still doesn't work is the math correct the formula i mean i don't even know if i am even doing the problem right
 
  • #10
It looks fine to me.
 
  • #11
Dick said:
It looks fine to me.

I agree. Ask the prof, and please post the final answer back here. Thanks.
 
  • #12
berkeman said:
I agree. Ask the prof, and please post the final answer back here. Thanks.

Well i he did a similar problem but i still don't get it

this is what he did y = x^2 -c^2

y = (0)^2-C^2

y = c^2-X^2

y = c^2-(0)
y = c^2

c^2-X^2=x^2-C^2

2c^2 = 2x^2

+-c = X


area integral a to b (y top - y bottom)dx


110 = integral -c to c (c^2-x^2)-(x^2-c^2)
 
  • #13
I think he is figuring out which curve is on top of the region (y=c^2-x^2) and which is on the bottom (y=x^2-c^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.
 
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  • #14
Dick said:
I think he is figuring out which curve is on top of the region (y=c^2-x^2) and which is on the bottom (y=x^2-c^2) and then the x boundaries +c and -c. He think 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.

thx its right
 
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