A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

Find the postive value of c such that the area of the the region enclosed by the parabolas y = x^2 -c^2 and y = c^2 -x^2 is 72.

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I thought I had to show c^2=x^2 so x=+-c so I evaluate from -c to c. But the equation is throwing me off if I was even right.

  2. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah that's right so far, let me make sure I have the answer though...

  3. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok, we got the top graph is c^2-x^2, and the bottom is x^2-c^2, so our integral is\[\int_{-c}^{c}c^2-x^2-(x^2-c^2)dx=\int_{-c}^{c}2c^2-2x^2dx\]the integrand is even so we can simplify this a bit\[=2\int_{0}^{c}2c^2-2x^2=2(2c^2x-\frac23x^3)|_{0}^{c}=2(2c^3-\frac23c^3)=72\]

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh I was doing it the other way. I put f(x) to be (x^2-c^2) and g(x) to be the other one and was getting a werid answer.

  5. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, it's a really good idea to sketch these thing out when you're not 100% on them. many people neglect that, and it costs them. welcome!

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Speaking of sketching, can you check what I may have done wrong here. It says y=x^2 and y=8-2x find area between x-axis and 2 curves. So I did \[\int\limits_{-4}^{2}x^2-(8-2x)\] it gives me a negative area?

  7. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah it sure does... think about this: what is x^2 when x=0 ? what is 8-2x when x=0 ?

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So wait did I do it wrong? I thought I should find where they intercept and then take the intergral. Iver never done one like this with the "between x-axis" part so not sure.

  9. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, but just answer my question let f(x)=x^2 let g(x)=8-2x what is f(0) ? what is g(0) ? picking a point in your interval of integration to check will tell you which graph should be on top and which should be on the bottom

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well x^2 would be 0 and 8-2(0) would be 4

  11. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so 8-2x should be on the top, yes?

  12. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh so it should be (8-2x)-(x2) instead?

  13. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yep 8-2x>x^2 on the interval you are talking about http://www.wolframalpha.com/input/?i=plot+x%5E2%2C8-2x

  14. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ...hence the negative answer

  15. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay let me try this and see if it is better

  16. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    should be :D

  17. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    dang I wish I knew all this like you ha would make this extensive homework easier

  18. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Just practice is all. What I study is still hard for me 'cause it's new. By the time you learn it it's on to the next thing :P

  19. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There is only one more question I had i doubt i can explain it here though. Let me try quick.

  20. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When you are finding the area of 2 lines with 3 shaded regions, how do you find the middle one when points arent given? the lines are y=x/3 and y=x^3/3

  21. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I can get the first region because i just evaluate from -2 to 0. The rest never seen before since some is above the x/3 line and some before

  22. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I only see two shaded regions between these graphs: between -1,0 and 0,1

  23. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Our drawing is from -2 and 0, 0 to 3 ish? but some is above the x/3 line and some below

  24. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac x3=\frac{x^3}3\to x=x^3\to x(1-x^2)=0\to x=\left\{ -1,0,1 \right\}\]are the intervals here's a visual http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3

  25. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1328632999874:dw|

  26. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not the best drawing but it wants the regions

  27. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    click the link for a better image of the graph

  28. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    does it tell you it wants the region between the graphs, or does it give you a specific interval to integrate?

  29. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The shaded region I drew. I think thats the region between graphs?

  30. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but the last region you drew on the right is unbounded: it goes to infinity again click the link I posted, there are only 2 regions enclosed between the graphs: http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3

  31. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well i think its bounded between the line and x=3

  32. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so he wants the area from the line at x=3 so it now has bounds i guess

  33. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then your intervals are\[[-1,0][0,1][1,3]\]

  34. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so you need to make sure that in each interval you have the proper graph on top, and proper graph on bottom

  35. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are u sure its not -2,0?

  36. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my book has the points (-2,2/3) as one point and (3,6) as another

  37. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if you want the region between the graphs then\[\frac{-2}{3}\neq\frac{(-2)^3}3\]so no

  38. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    at least that's not a point of intersection

  39. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (-2,-2/3)

  40. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    waitt

  41. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if you are told to integrate from x=-2 to x=3 then so be it, but those are not intersection points...

  42. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i feel stupid sorry. the line is x^3/3 -x

  43. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and x/3

  44. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ah... it happens

  45. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    should make more sense

  46. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{x^3}3-x=\frac x3\to\frac{x^3}3=\frac{4x}{3}\to x(x^2-4)=0\]\[x=\left\{ -2,0,2 \right\}\]so it's still not the right interval :/

  47. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well the 3rd shaded region the lines dont intercept

  48. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Its just from the point of (3,6) straight down to (3,1)

  49. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok, so we have from\[[-2,0][0,2][2,3]\]look at the graph http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3-x on the first interval x^3/3-x is on top on the second, x/3 is on top on the third, x^3/3 is on top again

  50. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have to go eat lunch, thanks for the help though. I will run it past my professor and maybe will get help before its due. Thanks

  51. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    me too, got class you should be able to set up the integral now though, good luck!

  52. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks, good luck

  53. myname
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is c a constant or a variable?

  54. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.