Question

Find the area of the region that lies inside the first curve and outside the second curve.
r^2 = 8 cos 2theta , r = 2
Ok so this is what i came up with, but its still incorrect, i'm missing something..
\[1/2 \int\limits_{-\pi/6}^{\pi/6} 8\cos 2\theta - 4~ d \theta \]

Answer 1

i have absolutely no idea sorry

Answer 2

I think there might be a mistake in the limits. I am trying to graph it .

Answer 3

don't you need to setup a double integral for this?

Answer 4

it is not necessary to

Answer 5

baru

not if you just plug and play \(A = \int \dfrac{1}{2} r^2 \; d \theta\)

the limits are just wrong

Answer 6

$$\large{
\int_{\alpha }^{\beta } \frac 1 2 ( r_o^2 - r_i^2 ) ~d\theta
\\
2\int_{-\pi /6 }^{\pi/6 } \frac 1 2 ( r_o^2 - r_i^2 ) ~d\theta
\\
2\int_{-\pi /6 }^{\pi/6 } \frac 1 2 ( 8 \cos ( 2 \theta ) -2^2 ) ~d\theta

}
$$

Answer 7

the \(-\pi/6 \to \pi /6\) gives half the area

so \[\dfrac{A}{\color{red}{2}} = 1/2 \int\limits_{-\pi/6}^{\pi/6} 8\cos 2\theta ~ d \theta \]

that's A **for the lemniscate**

then for the circle, the area to be deducted from A is \(2 \times \pi 2^2 \times \dfrac{\pi /3}{2 \pi}\), no need to integrate

Answer 8

Looks like you set it up right, but neglected the left half region.

Answer 9

ohh... i see it now
the same thing anyway follows from the double integral
\[\iint rdrd \theta=\int\limits \frac{r^2}{2}d \theta\]

Answer 10

The final answer I obtained was
$$ 4\sqrt{3}-(4/3) \pi$$

Answer 11

owen

agree

Answer 12

thank you! i wanted to know why it was half the area (i looked at the solutions manual for the exact value and i got half of that)
but how would i know determine the correct limits of integration then?

Answer 13

oh! okay i see why we double it now.

Answer 14

thanks man