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