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Finding area of shaded region

by intelli
Tags: region, shaded
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intelli
#1
Jun10-09, 05:38 PM
P: 20
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 dont understand how to do this problem any help would be greatly appreciated
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berkeman
#2
Jun10-09, 05:51 PM
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Quote Quote by intelli View Post
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 dont understand how to do this problem any help would be greatly appreciated
How did you get 3.49? Show us your work please.
Dick
#3
Jun10-09, 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.

intelli
#4
Jun10-09, 06:01 PM
P: 20
Finding area of shaded region

Quote Quote by berkeman View Post
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
intelli
#5
Jun10-09, 06:01 PM
P: 20
Quote Quote by berkeman View Post
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
Dick
#6
Jun10-09, 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?
intelli
#7
Jun10-09, 07:18 PM
P: 20
Quote Quote by Dick View Post
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 doesnt work
berkeman
#8
Jun10-09, 07:33 PM
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Quote Quote by intelli View Post
no i tried that it doesnt work
I get 3.4552116... That's not quite 3.45 to 3 sig figs.
intelli
#9
Jun10-09, 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
Dick
#10
Jun10-09, 09:25 PM
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It looks fine to me.
berkeman
#11
Jun10-09, 10:38 PM
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Quote Quote by Dick View Post
It looks fine to me.
I agree. Ask the prof, and please post the final answer back here. Thanks.
intelli
#12
Jun11-09, 03:47 PM
P: 20
Quote Quote by berkeman View Post
I agree. Ask the prof, and please post the final answer back here. Thanks.
Well i he did a similar problem but i still dont 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)
Dick
#13
Jun11-09, 03:57 PM
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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.
intelli
#14
Jun11-09, 04:10 PM
P: 20
Quote Quote by Dick View Post
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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