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Pippa

  • 2 years ago

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

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  1. Pippa
    • 2 years ago
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    |dw:1327896377720:dw|

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

  3. Pippa
    • 2 years ago
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    Hey ishaan

  4. Pippa
    • 2 years ago
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    It is a circle

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

  6. shaan_iitk
    • 2 years ago
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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?

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

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

  9. Pippa
    • 2 years ago
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    no

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

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

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

  13. Jemurray3
    • 2 years ago
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    lengthy*

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

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

  16. rivermaker
    • 2 years ago
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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 \[\int\limits_{a}^{b} f(x) dx\]

  17. Pippa
    • 2 years ago
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    okkk

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

  19. Jemurray3
    • 2 years ago
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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?

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

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

  22. rivermaker
    • 2 years ago
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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

  23. rivermaker
    • 2 years ago
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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

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

  25. Jemurray3
    • 2 years ago
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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.

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

  27. Pippa
    • 2 years ago
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    Thanks guys

  28. shaan_iitk
    • 2 years ago
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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.

  29. Jemurray3
    • 2 years ago
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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.

  30. Pippa
    • 2 years ago
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    shaan=Ishaan

  31. shaan_iitk
    • 2 years ago
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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.

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

  33. Jemurray3
    • 2 years ago
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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 :)

  34. Pippa
    • 2 years ago
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    hehe Jeymurray is correct

  35. shaan_iitk
    • 2 years ago
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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..

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

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

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

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

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

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

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

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

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

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