Finding area of shaded region

  1. Jun 10, 2009 #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
     
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  3. Jun 10, 2009 #2

    berkeman

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    How did you get 3.49? Show us your work please.
     
  4. Jun 10, 2009 #3

    Dick

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    It's hard to tell what you did wrong if you don't show us how you got 3.49.
     
  5. Jun 10, 2009 #4
    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. Jun 10, 2009 #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
     
  7. Jun 10, 2009 #6

    Dick

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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?
     
  8. Jun 10, 2009 #7
    no i tried that it doesnt work
     
  9. Jun 10, 2009 #8

    berkeman

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    I get 3.4552116... That's not quite 3.45 to 3 sig figs.
     
  10. Jun 10, 2009 #9
    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. Jun 10, 2009 #10

    Dick

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    It looks fine to me.
     
  12. Jun 10, 2009 #11

    berkeman

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    I agree. Ask the prof, and please post the final answer back here. Thanks.
     
  13. Jun 11, 2009 #12
    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. Jun 11, 2009 #13

    Dick

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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.
     
    Last edited: Jun 11, 2009
  15. Jun 11, 2009 #14
    thx its right
     
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