Question

Hi everyone! Can someone please tell me if I set up my integral correctly? I have to find the area of the inner circle. I will post my formula and answer, if someone could confirm my work that would be great! :o)

Answer 1

r=2-4sin(theta) is the graph

Answer 2

my integral is as follows: \[1/2\int\limits_{\pi/6}^{\pi/2}(2-4\sin \theta)^{2} d \theta \times 2\]

Answer 3

my answer was \[4\pi-6\sqrt{3}\]

Answer 4

When I graphed the polar curve from zero to pi/2, I basically got 1/2 of the inner circle, so I just multiplied the area formula times 2...can anyone confirm if I did this correctly? Thanks :o)

Answer 5

I will leave this question open but I have to go to sleep. If anyone can confirm or deny my answer I would appreciate it! Thanks! :o)

Answer 6

Why is your upper limit pi/2?

Answer 7

That would explain why you're only getting half of the inner loop.

To find out your limits you're going to have to find where the curve crosses the origin (at r=0)

\[r=0=2-4 \sin (\theta)\]
\[\sin(\theta)=\frac{1}{2}\]

So your limits will be
\[\theta = \frac{\pi}{6},\frac{5\pi}{6}\]

Answer 8

hi...I will explain why I used pi/2...

Answer 9

But your answer is correct

Answer 10

i physically graphed the curve...when theta = zero, r = 2 so your first point is (2,0)...
when theta is pi/2, r = -2...
the other two angles pi/4 and pi/3 simply lead you tpwards the point at angle pi/2...
if you look at the entire graph, going from angle zero to pi/2 looks like it draws exactly half of the inner circle...does that make sense? hold on and i will post a picture of the graph...

Answer 11

Yea that makes sense. You got the correct answer lol

Answer 12

Take a look at that graph...if it is wrong, can you please tell me where I messed up? :o)

Answer 13

I would guess there would be a couple other ways of finding the area, but this was the first way I tried

Answer 14

Yea, it's correct :)

Answer 15

yay! :o)

Answer 16

I have a conceptual question for you if you don't mind...

Answer 17

Go for it :)

Answer 18

if i'm not mistaken, I could have also integrated between pi/6 and pi, that way I would have had to multiply by 2...
additionally, if you wanted to be really crazy, couldn't i have integrated from zero to 2pi and then subtracted the integral from pi/6 to pi right? :o)~ yikes!

Answer 19

i mean NOT multiply by 2

Answer 20

or heck, i think i could have even went from zero to 2pi then subtracted the (integral from pio/6 to pi/2 multiplied by 2)
lol...right?

Answer 21

Integrating from pi/6 to pi would add a little bit too much to your area