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anonymous
 4 years ago
Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?
anonymous
 4 years ago
Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[r_1=A\sin\theta\quad\Rightarrow\quad x^2+\left(y\frac{A}{2}\right)^2=\left(\frac{A}{2}\right)^2\]\[r_2=B\sin\theta\quad\Rightarrow\quad x^2+\left(y\frac{B}{2}\right)^2=\left(\frac{B}{2}\right)^2\]\[S=\frac{\pi}{4}\cdot\left(\min{\left\{A,B\right\}}\right)^2\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.0should be 3pi/2 , not sure how you got your answer

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 + (yB/2)^2 = (B/2)^2 directly integrating polar is much easier, btw so now you have two cartesian equations , and you did ... ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0im not sure how to to integrate it using cartesian equations, hmmm

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok i guess it can be done, but it is tricky

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0These are circles. Why do you integrate?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0you want the area inside bounded by the two circles, you changed to cartesian so i integrated cartesian. the polar integration is much easier

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i am not sure why you changed to cartesian, but either way works

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perl,can you draw these circles?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Which does region lie inside both circles?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0I guess i solved a related question

perl
 4 years ago
Best ResponseYou've already chosen the best response.0yours is still wrong though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0limits of integration are 0 and pi, not 0 and 2pi.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0so i calculated area twice?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Between pi and 2pi circles do not exist.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok so you did integrate using polar, good , so you were just using the cartesian as an aid to graph it

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i set A sin theta = B sin theta

perl
 4 years ago
Best ResponseYou've already chosen the best response.0why dont circles exist between pi and 2pi?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because r cannot be negative

perl
 4 years ago
Best ResponseYou've already chosen the best response.0yes it can in polar coordinates

perl
 4 years ago
Best ResponseYou've already chosen the best response.0errr (r, theta ) = (r, theta + pi)

perl
 4 years ago
Best ResponseYou've already chosen the best response.0it just keeps going round and round in circles :)

perl
 4 years ago
Best ResponseYou've already chosen the best response.0but clearly i went twice around

perl
 4 years ago
Best ResponseYou've already chosen the best response.0that is why I got pi instead of pi/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[r=\sqrt{x^2+y^2}\ge0\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.0in polar we allow negative r

perl
 4 years ago
Best ResponseYou've already chosen the best response.0well 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
 4 years ago
Best ResponseYou've already chosen the best response.0anyways, we define (r, theta +pi) = (r,theta) , so its not a problem.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For this problem, your solution is very complicated and unnecessary.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0which one? i was solving a different problem

perl
 4 years ago
Best ResponseYou've already chosen the best response.0finding the region bounded by the two semicircles, i think its called the 'lune'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0these are not semicircles

perl
 4 years ago
Best ResponseYou've already chosen the best response.0right, i solved a different problem , the area bounded by the arcs . the lune

perl
 4 years ago
Best ResponseYou've already chosen the best response.0might not be the lune either not sure what you would call it, the area bounded by the inner and outer circles?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0and i got  a^2  b^2) * pi/4 as the area of the region bounded by the inner and outer circles

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find the area of the region that lies inside both curve

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i have the answer now, thanks

perl
 4 years ago
Best ResponseYou've already chosen the best response.0but i answered a different question , which is interesting

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i see , you didnt do any integration. you just used the area formula, pi * radius ^2 . nice

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i Have a difficult question, if you can help me...

perl
 4 years ago
Best ResponseYou've already chosen the best response.0solve 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
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0as in , i would like to know the other five solutions, i believe 4 real solutions and 2 non real

perl
 4 years ago
Best ResponseYou've already chosen the best response.0wow, i wish i could do that. did you use a computer algebra system for help, like maple or matlab?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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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