Finding area of shaded region

  1. 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
     
  2. jcsd
  3. berkeman

    Staff: Mentor

    How did you get 3.49? Show us your work please.
     
  4. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    It's hard to tell what you did wrong if you don't show us how you got 3.49.
     
  5. 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
     
  6. 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
     
  7. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    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?
     
  8. no i tried that it doesnt work
     
  9. berkeman

    Staff: Mentor

    I get 3.4552116... That's not quite 3.45 to 3 sig figs.
     
  10. 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
     
  11. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    It looks fine to me.
     
  12. berkeman

    Staff: Mentor

    I agree. Ask the prof, and please post the final answer back here. Thanks.
     
  13. 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)
     
  14. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    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.
     
    Last edited: Jun 11, 2009
  15. thx its right
     
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