Question

parametric equations help

Answer 1

@electrokid i am sure you can help me

Answer 2

this is how the polar plots look like
\[
r_1=2\cos^2\theta\\r_2=4\cos^2\theta\\
\text{Area between the curves:}\\
A=4\int_{r=r_1}^{r_2}\int_{\theta=0}^{\pi/2}r\,dr\,d\theta\\
A=4\left[r^2\over 2\right]_{2\cos^2\theta}^{4\cos^2\theta}\left[\theta\right]_0^{\pi/2}\\
A=?
\]

Answer 3

wait.. the integration..
\[
A=4\int_{\theta=0}^{\pi/2}\left[r^2\over 2\right]_{2\cos^2\theta}^{4\cos^2\theta}d\theta
\]

Answer 4

nice...wow

Answer 5

so if i integrate the stuff am i gonna get the area?

Answer 6

yep.

Answer 7

did you notice that I am integrating only in the first quadrant and multiplying it by "4"

Answer 8

ya why is that 4 for?

Answer 9

there are four quadrants in a 2D space!!

Answer 10

oh ya got it? and i am confused in integration...you gave me 2 integration, which one should i use

Answer 11

the second one.
in the first one, i made a mistake

Answer 12

error in the second step of the first integration post. I corrected that in my next post

Answer 13

ohk so is it asking me to integrate r^2/2 and substitute 4cos^2 theta - 2cos^2 theta but then whats the purpose of 0 to pi/2

Answer 14

it is a double integral. First integrating with "r", we get
\[
A=2\int_0^{\pi/2}\left[16\cos^4\theta-4\cos^4\theta\right]d\theta
\]now, we solve this integral

Answer 15

oh i know how to solve thanks....and where should we located so it wouldn't matter which company should we use

Answer 16

I'd say, anywhere within the range of the first one..\(r=2\cos^2\theta\)

Answer 17

why

Answer 18

because, that region is served by BOTH the antennae

Answer 19

got it thanks...would you help me with 2 more this kinda problems?

Answer 20

post em as new questions. will see