finding area of shaded region


by intelli
Tags: region, shaded
intelli
intelli is offline
#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
Phys.Org News Partner Science news on Phys.org
Simplicity is key to co-operative robots
Chemical vapor deposition used to grow atomic layer materials on top of each other
Earliest ancestor of land herbivores discovered
berkeman
berkeman is offline
#2
Jun10-09, 05:51 PM
Mentor
berkeman's Avatar
P: 39,595
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
Dick is offline
#3
Jun10-09, 05:52 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
It's hard to tell what you did wrong if you don't show us how you got 3.49.

intelli
intelli is offline
#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
intelli is offline
#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
Dick is offline
#6
Jun10-09, 06:21 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
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
intelli is offline
#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
berkeman is offline
#8
Jun10-09, 07:33 PM
Mentor
berkeman's Avatar
P: 39,595
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
intelli is offline
#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
Dick is offline
#10
Jun10-09, 09:25 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
It looks fine to me.
berkeman
berkeman is offline
#11
Jun10-09, 10:38 PM
Mentor
berkeman's Avatar
P: 39,595
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
intelli is offline
#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
Dick is offline
#13
Jun11-09, 03:57 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
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
intelli is offline
#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


Register to reply

Related Discussions
Find n for the shaded region Calculus & Beyond Homework 6
Finding Area of Region Calculus & Beyond Homework 6
Area of region R??? Calculus & Beyond Homework 9
Pre-Calculus - Regarding finding the area of a certain region Introductory Physics Homework 4
finding the area of the region Introductory Physics Homework 4