# Finding area of shaded region

by intelli
 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
Mentor
P: 41,303
 Quote by intelli 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
 Sci Advisor HW Helper Thanks P: 25,235 It's hard to tell what you did wrong if you don't show us how you got 3.49.
P: 20

 Quote by berkeman 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
P: 20
 Quote by berkeman 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
 Sci Advisor HW Helper Thanks P: 25,235 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?
P: 20
 Quote by Dick 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
Mentor
P: 41,303
 Quote by intelli no i tried that it doesnt work
I get 3.4552116... That's not quite 3.45 to 3 sig figs.
 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
 Sci Advisor HW Helper Thanks P: 25,235 It looks fine to me.
Mentor
P: 41,303
 Quote by Dick It looks fine to me.
P: 20
 Quote by berkeman 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)
 Sci Advisor HW Helper Thanks P: 25,235 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.
P: 20
 Quote by Dick 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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