anonymous
  • anonymous
Can someone explain how we use an integral to find the area?
Mathematics
katieb
  • katieb
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katieb
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anonymous
  • anonymous
|dw:1327896377720:dw|
anonymous
  • anonymous
This is a circle or a sphere?
anonymous
  • anonymous
Hey ishaan

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anonymous
  • anonymous
It is a circle
anonymous
  • anonymous
I knwo the steps but I dont get what i am doing
anonymous
  • anonymous
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?
anonymous
  • anonymous
nooo just show some arrows and a h
anonymous
  • anonymous
Have you discussed polar integration yet? I doubt it, but just to be sure.... -.-
anonymous
  • anonymous
no
anonymous
  • anonymous
like the book used pythagorean theroem in the begginning. in order to get the equation of the integral
anonymous
  • anonymous
I guess everybody fell asleep at the comp like me -_-
anonymous
  • anonymous
No, but it seems rivermaker is typing a very length reply, so I was waiting to see ;)
anonymous
  • anonymous
lengthy*
anonymous
  • anonymous
hehe I was just teasing :D
anonymous
  • anonymous
river are you still going, or is it saying you're typing by mistake?? haha
anonymous
  • anonymous
|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 \[\int\limits_{a}^{b} f(x) dx\]
anonymous
  • anonymous
okkk
anonymous
  • anonymous
Mmm. That's kinda true, but not in this instance, unfortunately. It doesn't work quite like that.
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes i know trig subsitutions
anonymous
  • anonymous
I am in calc 2
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
Thanks rivermaker for ur time XD
anonymous
  • anonymous
@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.
anonymous
  • anonymous
ya ok will do :D
anonymous
  • anonymous
Thanks guys
anonymous
  • anonymous
@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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
shaan=Ishaan
anonymous
  • anonymous
@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.
anonymous
  • anonymous
nopes I am not Ishaan
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
hehe Jeymurray is correct
anonymous
  • anonymous
@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..
anonymous
  • anonymous
I havent learnt this method yet
anonymous
  • anonymous
Its basic school stuff .. have you had any exposure to calculus?
anonymous
  • anonymous
hehe idk I am taking skewl online
anonymous
  • anonymous
so like I am missing basic knowledge
anonymous
  • anonymous
Trying my hardest to teach everything to myself
anonymous
  • anonymous
At some points i just snap. LIke now my brain is fried
anonymous
  • anonymous
Thanks for ur help I really appreciate it
anonymous
  • anonymous
lol I get it now hehe
anonymous
  • anonymous
I am so excited :D

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