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Pippa
 3 years ago
Can someone explain how we use an integral to find the area?
Pippa
 3 years ago
Can someone explain how we use an integral to find the area?

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shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0This is a circle or a sphere?

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1I knwo the steps but I dont get what i am doing

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0Where 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?

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1nooo just show some arrows and a h

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Have you discussed polar integration yet? I doubt it, but just to be sure.... .

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1like the book used pythagorean theroem in the begginning. in order to get the equation of the integral

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1I guess everybody fell asleep at the comp like me _

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0No, but it seems rivermaker is typing a very length reply, so I was waiting to see ;)

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1hehe I was just teasing :D

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0river are you still going, or is it saying you're typing by mistake?? haha

rivermaker
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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 coordinates 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 \[\int\limits_{a}^{b} f(x) dx\]

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Mmm. That's kinda true, but not in this instance, unfortunately. It doesn't work quite like that.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Integrating 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?

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1yes i know trig subsitutions

rivermaker
 3 years ago
Best ResponseYou've already chosen the best response.1For 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

rivermaker
 3 years ago
Best ResponseYou've already chosen the best response.1Since 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

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks rivermaker for ur time XD

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0shaan, 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.

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0nopes I am not Ishaan

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 :)

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1hehe Jeymurray is correct

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0@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..

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1I havent learnt this method yet

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0Its basic school stuff .. have you had any exposure to calculus?

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1hehe idk I am taking skewl online

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1so like I am missing basic knowledge

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1Trying my hardest to teach everything to myself

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1At some points i just snap. LIke now my brain is fried

Pippa
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks for ur help I really appreciate it
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