anonymous
  • anonymous
Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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nikvist
  • nikvist
\[r_1=A\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{A}{2}\right)^2=\left(\frac{A}{2}\right)^2\]\[r_2=B\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{B}{2}\right)^2=\left(\frac{B}{2}\right)^2\]\[S=\frac{\pi}{4}\cdot\left(\min{\left\{A,B\right\}}\right)^2\]
perl
  • perl
this does not work
perl
  • perl
try A = 2 , B = 1

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perl
  • perl
should be 3pi/2 , not sure how you got your answer
perl
  • perl
ok looks like you did a polar conversion r = A sin theta r^2 = r* A sin theta x^2 + y^2 = A r* sin theta x^2 + y^2 = Ay x^2 + y^2 - Ay = 0 complete square x^2 + (y -A/2)^2 = (A/2)^2 similiarly x^2 + (y-B/2)^2 = (B/2)^2 directly integrating polar is much easier, btw so now you have two cartesian equations , and you did ... ?
perl
  • perl
im not sure how to to integrate it using cartesian equations, hmmm
perl
  • perl
ok i guess it can be done, but it is tricky
perl
  • perl
wlog A > B > 0 so
nikvist
  • nikvist
These are circles. Why do you integrate?
perl
  • perl
|dw:1328800285296:dw|
perl
  • perl
you want the area inside bounded by the two circles, you changed to cartesian so i integrated cartesian. the polar integration is much easier
perl
  • perl
|dw:1328800478524:dw|
perl
  • perl
i am not sure why you changed to cartesian, but either way works
nikvist
  • nikvist
Perl,can you draw these circles?
perl
  • perl
|dw:1328800689137:dw|
nikvist
  • nikvist
Which does region lie inside both circles?
perl
  • perl
lol
perl
  • perl
I guess i solved a related question
perl
  • perl
yours is still wrong though
perl
  • perl
|dw:1328801098170:dw|
perl
  • perl
|dw:1328801200075:dw|
nikvist
  • nikvist
limits of integration are 0 and pi, not 0 and 2pi.
perl
  • perl
ok
perl
  • perl
ok i agree now
perl
  • perl
so i calculated area twice?
nikvist
  • nikvist
Between pi and 2pi circles do not exist.
perl
  • perl
ok so you did integrate using polar, good , so you were just using the cartesian as an aid to graph it
perl
  • perl
i set A sin theta = B sin theta
perl
  • perl
solutions are pi*k
perl
  • perl
why dont circles exist between pi and 2pi?
nikvist
  • nikvist
because r cannot be negative
perl
  • perl
yes it can in polar coordinates
perl
  • perl
errr (r, theta ) = (-r, theta + pi)
perl
  • perl
it just keeps going round and round in circles :)
perl
  • perl
but clearly i went twice around
perl
  • perl
that is why I got pi instead of pi/2
nikvist
  • nikvist
\[r=\sqrt{x^2+y^2}\ge0\]
perl
  • perl
in polar we allow negative r
perl
  • perl
well if you have a TI 84 or graphing calculator, you can graph r = sin theta , and watch what happens when theta > pi . i think
perl
  • perl
anyways, we define (-r, theta +pi) = (r,theta) , so its not a problem.
perl
  • perl
for instance, suppose theta = 3pi/2, and given r(theta) = sin theta . we have ( r, theta ) = ( sin 3pi/2, 3pi/2) = (-1, 3pi/2) = (1 , 3pi/2 + - pi) = ( 1, pi/2) , so we can make any negative radius positive, using that trick
nikvist
  • nikvist
For this problem, your solution is very complicated and unnecessary.
perl
  • perl
which one? i was solving a different problem
perl
  • perl
finding the region bounded by the two semicircles, i think its called the 'lune'
nikvist
  • nikvist
these are not semicircles
perl
  • perl
right, i solved a different problem , the area bounded by the arcs . the lune
perl
  • perl
might not be the lune either not sure what you would call it, the area bounded by the inner and outer circles?
perl
  • perl
and i got | a^2 - b^2) |* pi/4 as the area of the region bounded by the inner and outer circles
nikvist
  • nikvist
Find the area of the region that lies inside both curve
perl
  • perl
yes i misread it
perl
  • perl
i have the answer now, thanks
perl
  • perl
but i answered a different question , which is interesting
perl
  • perl
i see , you didnt do any integration. you just used the area formula, pi * radius ^2 . nice
perl
  • perl
i Have a difficult question, if you can help me...
perl
  • perl
its elementary though
perl
  • perl
Can i post it here?
perl
  • perl
solve for all real and non real the system of equations. xy - z/3 = xyz +1 yz - x/3 = xyz -1 xz - y/3 = xyz - 9
perl
  • perl
if you have a solution, please email riemanngalois@gmail.com I already have the integer solution (3,0,-3) and i know that you can plug it into wolfram. I want to know how did wolfram solve this. thanks
perl
  • perl
as in , i would like to know the other five solutions, i believe 4 real solutions and 2 non real
nikvist
  • nikvist
brute force solution
1 Attachment
perl
  • perl
wow, i wish i could do that. did you use a computer algebra system for help, like maple or matlab?
perl
  • perl
yes so we have to test the solutions, to see if any are extraneous. looks good so you did equation 1 - equation 2 , then you solved equation 3 . the logic of this gets me a bit confused. you got the quadratic of z, treating y as a constant. then z = f(y). so then x = g(y) back substituting. so shouldnt it be ( g(y), y, f(y) ) , infinite solutions>?
perl
  • perl
ok so you did (1) - (2), (3) -> you got z = f(y) , x = g(y) . you can then plug these into any of the three equations? does it matter?

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