Question

Find the area of the region.
Inside r = 2a cos(θ) and outside r = a

Answer 1

first find the points of intersections

solve
a=2acos(theta)
let me know what you got.

Answer 2

hmmm

Answer 3

how to find poi?

Answer 4

solve this equation
\[\Large a=2acos(\theta)\]

Answer 5

solve for theta? a/a = 1 = 2cos(theta) arccos(1/2) = theta?

Answer 6

yes correct

Answer 7

now what do I do?

Answer 8

or is that the answer........?

Answer 9

take your calculator and find arccos(1/2)

Answer 10

pi/3

Answer 11

another value as well.
because the graph is intersecting in 4th Quadrant as well.

Answer 12

5pi/3?

Answer 13

do you think 2pi/3 is in 4th quadrant :P

Answer 14

I'm really bad with this stuff sorry.. Some of my basic math was destroyed by horrid teachers...

Answer 15

ok the other angle is
\[\Large 2\pi-\pi/3=5\pi/3\] sorry i did not look above you had found it it.

Answer 16

:)

Answer 17

ok now the range is in our hand s up integral
\[\Large Area=\int\limits_{a}^{b}\frac{1}{2}r^2d \theta\]

Answer 18

int from pi/3 to 5pi/3 of 1/2 r^2d theta huh?

Answer 19

which would equal r^3/6 from those points :P.

Answer 20

\[\Large \int\limits_{\frac{5\pi}{3}}^{\frac{\pi}{3}}\frac{1}{2}(2acos(\theta))^2-a^2)d \theta\]
can you do this now ?

Answer 21

5pi/3 is on the bottom.........?

Answer 22

@sami-21

Answer 23

also how do we use the a's in this? Confused stilll :(.

Answer 24

yes it is in the bottom.
also treat a as a constant.

Answer 25

I mean we solved for it, but you want me to use it in that form?

Answer 26

use it in that form . just ntegrate the above integral and you have the required area.

Answer 27

well can we cancel them? It looks like we can get rid of the 1/2 with the 2 in front.

Answer 28

2 is under the square (2acos(theta))^2

Answer 29

true.......... hmm....

Answer 30

you're missing an open paren up there btw....

Answer 31

Maybe that'sd why I'm so konfused....

Answer 32

yes i am missing one paren. :P

Answer 33

Idk if I've ever seen an integral with 2 variables...

Answer 34

theta would be x.... but you said a is a "constant?" I mean but it could be anything... right?

Answer 35

:O

Answer 36

@sami-21 done?

Answer 37

there is only obe variable thetha.
let me explain little bit.

Answer 38

ok ty :)

Answer 39

\[\Large \frac{1}{2} \int\limits_{\frac{5\pi}{3}}^{\frac{\pi}{3}}[4a^2\cos^2(\theta)-a^2)d \theta\]
or
\[\Large \frac{a^2}{2} \int\limits\limits_{\frac{5\pi}{3}}^{\frac{\pi}{3}}[(4\cos^2(\theta)-1)d \theta]\]

i hope now you can do it.

Answer 40

hmm

Answer 41

:)

Answer 42

a^2/2[(pi/3+sin(2pi/3)) -(5pi/3 + sin(10pi/3))] ????

Answer 43

which = sqrt(3)-(4 π)/3

Answer 44

with a^2/2 also there.

Answer 45

you have missed second term
while integrating
there should be second term as well when you expand integral sign.
\[\Large \int\limits_{\frac{5\pi}{3}}^{\frac{\pi}{3}} 1d \theta\]

Answer 46

http://www.wolframalpha.com/input/?i=+integral+%284cos%5E2%28x%29-1%29+from+5pi%2F3+to+pi%2F3

Answer 47

:(?