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

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- 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

this does not work

- perl

try A = 2 , B = 1

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- perl

should be 3pi/2 , not sure how you got your answer

- 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

im not sure how to to integrate it using cartesian equations, hmmm

- perl

ok i guess it can be done, but it is tricky

- perl

wlog A > B > 0 so

- nikvist

These are circles. Why do you integrate?

- perl

|dw:1328800285296:dw|

- 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

|dw:1328800478524:dw|

- perl

i am not sure why you changed to cartesian, but either way works

- nikvist

Perl,can you draw these circles?

- perl

|dw:1328800689137:dw|

- nikvist

Which does region lie inside both circles?

- perl

lol

- perl

I guess i solved a related question

- perl

yours is still wrong though

- perl

|dw:1328801098170:dw|

- perl

|dw:1328801200075:dw|

- nikvist

limits of integration are 0 and pi, not 0 and 2pi.

- perl

ok

- perl

ok i agree now

- perl

so i calculated area twice?

- nikvist

Between pi and 2pi circles do not exist.

- perl

ok so you did integrate using polar, good , so you were just using the cartesian as an aid to graph it

- perl

i set A sin theta = B sin theta

- perl

solutions are pi*k

- perl

why dont circles exist between pi and 2pi?

- nikvist

because r cannot be negative

- perl

yes it can in polar coordinates

- perl

errr
(r, theta ) = (-r, theta + pi)

- perl

it just keeps going round and round in circles :)

- perl

but clearly i went twice around

- perl

that is why I got pi instead of pi/2

- nikvist

\[r=\sqrt{x^2+y^2}\ge0\]

- perl

in polar we allow negative r

- 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

anyways, we define (-r, theta +pi) = (r,theta) , so its not a problem.

- 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

For this problem, your solution is very complicated and unnecessary.

- perl

which one? i was solving a different problem

- perl

finding the region bounded by the two semicircles, i think its called the 'lune'

- nikvist

these are not semicircles

- perl

right, i solved a different problem , the area bounded by the arcs . the lune

- perl

might not be the lune either
not sure what you would call it, the area bounded by the inner and outer circles?

- perl

and i got | a^2 - b^2) |* pi/4 as the area of the region bounded by the inner and outer circles

- nikvist

Find the area of the region that lies inside both curve

- perl

yes i misread it

- perl

i have the answer now, thanks

- perl

but i answered a different question , which is interesting

- perl

i see , you didnt do any integration. you just used the area formula, pi * radius ^2 . nice

- perl

i Have a difficult question, if you can help me...

- perl

its elementary though

- perl

Can i post it here?

- 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

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

as in , i would like to know the other five solutions, i believe 4 real solutions and 2 non real

- nikvist

brute force solution

##### 1 Attachment

- perl

wow, i wish i could do that. did you use a computer algebra system for help, like maple or matlab?

- 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

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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