Question

find the points of intersection

Answer 1

set them equal, then solve for sin(theta) , then take inverse sin to get angle

Answer 2

correct

Answer 3

Okay. Do you knkow how i would set up an integral to find the area of their intersection?

Answer 4

hmm not great at integrating polar equations...lets ask someone else
@Hero
@UnkleRhaukus
@jim_thompson5910

Answer 5

my instinct says it will be
\[\int\limits_{\pi/6}^{5\pi/6}2\sin \theta - (3/2-\sin \theta) d \theta\]

Answer 6

Actually I think both of our answers are incorrect I am not sure. Hopefully we will hear from someone else

Answer 7

but i think the rules for polar equations are different...oh yeah there should be a 1/2 in there somewhere
sorry :{

Answer 8

Its alright thanks for the help on the first bit

Answer 9

I need help setting up the integral I will try to solve it from there

Answer 10

here's a good page on polar integration

http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx

Answer 11

oops made a typo

Use the formula

\[\Large A = \int_{a}^{b} \frac{1}{2}r^2 d\theta \]

In this case, you want to find the difference of the two areas to find the area of the region between the two polar graphs, so this means


\[\Large A = \int_{a}^{b} \frac{1}{2}\left(\left(r_{2}\right)^2 - \left(r_{1}\right)^2 \right)d\theta \]

Answer 12

that's one of the correct steps to get to your answer

Answer 13

Now I would finish out the integral and that would be the area?

Answer 14

yes

Answer 15

So I don't have to multiply by 2 or anything ?

Answer 16

what do you mean

Answer 17

Ive seen some questions where they change the bounds from -pi/2 to pi/2 to something like 0 to pi/2 times 2

Answer 18

oh, they're probably taking advantage of symmetry

Answer 19

if you know the top half area, and the figure is symmetrical along the x-axis, then the bottom half is the same ---> total area = 2*(top area)

Answer 20

So this would be symmetrical too

Answer 21

Sorry It would not

Answer 22

yes, you could go from one intersection to pi/2

Answer 23

you don't have to though

Answer 24

Would i just take the derivative which would be 2 cos theta and plug in pi/6

Answer 25

when theta = pi/6, r is

r = 2*sin(theta)

r = 2*sin(pi/6)

r = 2*(1/2)

r = 1

So we have r = 1 and theta = pi/6

Now use the formulas

x = r*cos(theta)

and

y = r*sin(theta)

to find x and y, this will convert the point (r,theta) to the cartesian point (x,y)

-----------------------------------

Now convert r = 2*sin(theta) to rectangular form

Use the equations x^2 + y^2 = r^2 and y = r*sin(theta)


r = 2*sin(theta)

r*r = r*2*sin(theta)

r^2 = 2*r*sin(theta)

r^2 = 2y

x^2 + y^2 = 2y

x^2 + y^2 - 2y = 0

x^2 + y^2 - 2y + 1 = 1

x^2 + (y - 1)^2 = 1

So the polar equation r = 2*sin(theta) converts to the rectangular equation x^2 + (y - 1)^2 = 1


Derive that last equation. Then find the slope of the line at the (x,y) point you found previously.

Answer 26

What is the previous point?

Answer 27

Whatever (r,theta) = (1,pi/6) converts to in rectangular (x,y) form

Answer 28

yes

Answer 29

as the slope?

Answer 30

one moment

Answer 31

yes I'm getting that as well

Answer 32

yw