Question

HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)

Answer 1

let me guess, third year calculus?

Answer 2

yeah why?

Answer 3

It was torture for me :) let me see if I can help, I need some scratch paper though

Answer 4

ok thanks man its so tough

Answer 5

i can relate, gimmie a sec

Answer 6

ok

Answer 7

have you done double integrals?

Answer 8

Not yet

Answer 9

dang, okay gotta think of how to do this using single...

Answer 10

we haven't gotten that far we are in chapter 9 with areas of regions and in curves

Answer 11

give me as much info as you can then

Answer 12

well the question is just right there lol, idk what else you want

Answer 13

we have done polar coordinates, taylor series , maclaurin series

Answer 14

okay good, you can't do this in rectangular, polar is your best, no only shot unless you want roots. it's also cleaner. I just need to remember how to do a single integral sadly. I haven't done Calc 3 in a long time

Answer 15

Yeah I hear you, I'm looking at this problem and I don't even know where to start, if you can show me step by step what you do, that would be so helpful

Answer 16

\[8\sin(2\theta)=8\sin(\theta)\]
\[\sin(2\theta)=\sin(\theta)\]
\[2\sin(\theta)\cos(\theta)=\sin(\theta)\]
\[2\sin(\theta)\cos(\theta)-\sin(\theta)=0\]
\[\sin(\theta)(2\cos(\theta)-1)=0\]
\[\sin(\theta)=0\] and \[2\cos(\theta)-1=0\]

Solve for theta and these are your limits of integration.

Answer 17

Notice how I cancelled the 8 but didn't cancel the sin(theta). You never want to cancel out anything that isn't a constant since it might be a potential solution.

Answer 18

ok I see

Answer 19

Can you pick it up from there since you've done polar coordinates? I don't know if you've done integration using polar yet or not.

Answer 20

so its 2 cos(theta) = 1,
so cos(theta) = 1/2?

Answer 21

But believe me, you'll need a strong grip on polar since double and triple integrals build on that. Cylindrical is like a second layer to polar and spherical...well just be strong in polar and double and triple will be easier for you

Answer 22

Yes

Answer 23

OK so i have sin(theta) = 0 and cos(theta) = 1/2

Answer 24

yep, and notice how both functions are even( you're in polar not rectangular) so you need to take into account the negative answer

Answer 25

Ok so now what?

Answer 26

what are your values for theta?

Answer 27

0 and 1/2?

Answer 28

come on, your values for theta, not sin(theta) and cos(theta). Think trig or geometry. This is calculus so if you're willing to take calc, especially calc 3, you should know how to take the inverse of a trig function.

Answer 29

I thought you meant those values though

Answer 30

theta refers to the angle. What are the angles? Stop and don't think calculus right now. Go back to your geometry or trig. How did you find an angle using trig functions like sine and cosine?

Answer 31

The circle, so wouldn't it be 30 degrees