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Pippa Group Title

Can someone explain how we use an integral to find the area?

  • 2 years ago
  • 2 years ago

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  1. Pippa Group Title
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    |dw:1327896377720:dw|

    • 2 years ago
  2. shaan_iitk Group Title
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    This is a circle or a sphere?

    • 2 years ago
  3. Pippa Group Title
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    Hey ishaan

    • 2 years ago
  4. Pippa Group Title
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    It is a circle

    • 2 years ago
  5. Pippa Group Title
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    I knwo the steps but I dont get what i am doing

    • 2 years ago
  6. shaan_iitk Group Title
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    Where is the trouble? ..also is any information given regarding this chord? .. what angle it suspends .. and is the width of the chord comparable to the radius?

    • 2 years ago
  7. Pippa Group Title
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    nooo just show some arrows and a h

    • 2 years ago
  8. Jemurray3 Group Title
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    Have you discussed polar integration yet? I doubt it, but just to be sure.... -.-

    • 2 years ago
  9. Pippa Group Title
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    no

    • 2 years ago
  10. Pippa Group Title
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    like the book used pythagorean theroem in the begginning. in order to get the equation of the integral

    • 2 years ago
  11. Pippa Group Title
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    I guess everybody fell asleep at the comp like me -_-

    • 2 years ago
  12. Jemurray3 Group Title
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    No, but it seems rivermaker is typing a very length reply, so I was waiting to see ;)

    • 2 years ago
  13. Jemurray3 Group Title
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    lengthy*

    • 2 years ago
  14. Pippa Group Title
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    hehe I was just teasing :D

    • 2 years ago
  15. Jemurray3 Group Title
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    river are you still going, or is it saying you're typing by mistake?? haha

    • 2 years ago
  16. rivermaker Group Title
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    |dw:1327897236799:dw| Let us take any function y = f(x) THe area under the curve y = f(x) is the sum of rectangular strips (like one shown near rhe middle of the picture) if you assume the co-ordinates of the strip as \[(x_{0}, 0), (x_{0}, y_{0}), (x_{1}, y_{1}), (x_{1},0) \] then the area of the strip is approximated by \[y_{0} \times (x_{1} - x_{0}) \] So the area under the curve is the sum of such strips \[\sum_{x=a}^{x=b}f(x) \Delta x, \] where \[\Delta x \] is the width of the strip at \[x\] Going to the limit as \[\Delta x \rightarrow 0, \] we get the area as \[\int\limits_{a}^{b} f(x) dx\]

    • 2 years ago
  17. Pippa Group Title
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    okkk

    • 2 years ago
  18. Jemurray3 Group Title
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    Mmm. That's kinda true, but not in this instance, unfortunately. It doesn't work quite like that.

    • 2 years ago
  19. Jemurray3 Group Title
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    Integrating over the area enclosed by a circle is a little bit more complicated for several reasons. Pippa, what level of calculus are you in, and what have you been learning recently? Trig substitutions, for instance?

    • 2 years ago
  20. Pippa Group Title
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    yes i know trig subsitutions

    • 2 years ago
  21. Pippa Group Title
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    I am in calc 2

    • 2 years ago
  22. rivermaker Group Title
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    For a circle use the fact that the area of the circle is 4 times the area of one quadrant; if you know the equation for a circle -- recast it in the form y = f(x) It should be a quadratic (degree 2) in x Integrating that should give you the answer as jemurray3 said some trig substitution will be needed to do the integration

    • 2 years ago
  23. rivermaker Group Title
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    Since I am from India, calc 2 does not exactly translate :-) I am trying to give just enough pointers so that you can still figure out and learn. DO ask for more details if needed

    • 2 years ago
  24. Pippa Group Title
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    Thanks rivermaker for ur time XD

    • 2 years ago
  25. Jemurray3 Group Title
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    @River, for reference, our Calculus II curriculum includes more advanced substitution methods, integrating volumes of rotation using techniques like the disk method, and basic work in infinite series, convergence, and improper integrals. @Pippa give that a shot, and if you need anything further then say so. The trig substitution is critical.

    • 2 years ago
  26. Pippa Group Title
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    ya ok will do :D

    • 2 years ago
  27. Pippa Group Title
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    Thanks guys

    • 2 years ago
  28. shaan_iitk Group Title
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    @pippa .. area of the segment you have mentioned would depend on how far the chord is from the center @jemurray and river: mathematics world over remains the same. Its not that if you in India then 2+2 becomes 5. Its upto individuals to make or break their skill.

    • 2 years ago
  29. Jemurray3 Group Title
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    shaan, in U.S. universities we have a standard calculus sequence, Calculus I, II, and III. I was just clarifying what material is covered in most Calculus II courses.

    • 2 years ago
  30. Pippa Group Title
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    shaan=Ishaan

    • 2 years ago
  31. shaan_iitk Group Title
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    @Jemurray3: I understand. We also have the same sequence of teaching. That is what I was trying to clarify. Its not USA or India or even say Iran. You go to IRAN and even there you would find course structure more or less similar. Its upto individuals how good they are. MIT doesn't make you intelligent, is the intelligent people who reach MIT.

    • 2 years ago
  32. shaan_iitk Group Title
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    nopes I am not Ishaan

    • 2 years ago
  33. Jemurray3 Group Title
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    I wouldn't necessarily say that, though. Not all smart people reach MIT or other top schools simply because they can only accept so many people. Furthermore, the teaching at MIT is in many cases far superior and more enriching than at other institutions. That's not to say you can't get a fantastic education anywhere else, but the top is the top for a reason :)

    • 2 years ago
  34. Pippa Group Title
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    hehe Jeymurray is correct

    • 2 years ago
  35. shaan_iitk Group Title
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    @Pippa .. you don't need polar integration for this.. Assume that the outer chord of the segment (the one with width delta h) suspends an angle (\[\theta + d \theta\] at the center. Also assume that the inner chord of the segment suspends an angle \[\theta\] . In this case you would have r*(d\[\theta\]) = h. Now area of any segment of a circle with angle \[\alpha\] is = r^2 *\[\alpha\]/(2pie) .. Area of the segment would be (area enclosed by the outer chord - area enclosed by the lower chord) = (r*\[\Delta h\])/(2*pie) .. I hope I am clear..

    • 2 years ago
  36. Pippa Group Title
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    I havent learnt this method yet

    • 2 years ago
  37. shaan_iitk Group Title
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    Its basic school stuff .. have you had any exposure to calculus?

    • 2 years ago
  38. Pippa Group Title
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    hehe idk I am taking skewl online

    • 2 years ago
  39. Pippa Group Title
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    so like I am missing basic knowledge

    • 2 years ago
  40. Pippa Group Title
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    Trying my hardest to teach everything to myself

    • 2 years ago
  41. Pippa Group Title
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    At some points i just snap. LIke now my brain is fried

    • 2 years ago
  42. Pippa Group Title
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    Thanks for ur help I really appreciate it

    • 2 years ago
  43. Pippa Group Title
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    lol I get it now hehe

    • 2 years ago
  44. Pippa Group Title
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    I am so excited :D

    • 2 years ago
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