# Math Help - find the area

1. ## find the area

of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?

2. Originally Posted by tomobson
of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?
You should find their intersections 1st.
Then find the area of each lines from one intersection to another.

3. Originally Posted by tomobson
of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?
Graphing this shows me that it is symmetric around the origin so I'll just calculate it in the first quadrant and multiply by two.

You first want to find where the two curves intersect. If $r^2=4\sin(2\theta)$ and $r^2=2$, these two curves intersect at $\theta=\frac{\pi}{12}$ and $\theta=\frac{5\pi}{12}$.

The formula you want to use is $\frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta$

Plugging in gives you $2\cdot\frac{1}{2}\int_{\pi/12}^{5\pi/12}(4\sin(2\theta)-2)\,d\theta$
Spoiler:
$= \bigg[-2\cos(2\theta)-2\theta\bigg]_{\pi/12}^{5\pi/12}$ $= \left[\left(\sqrt{3}-\frac{5\pi}{6}\right)-\left(-\sqrt{3}-\frac{\pi}{6}\right)\right] = 2\sqrt{3}-\frac{2\pi}{3}$

4. Originally Posted by redsoxfan325
Graphing this shows me that it is symmetric so I'll just calculate it in the first quadrant and multiply by four.

You first want to find where the two curves intersect. If $r^2=4\sin(2\theta)$ and $r^2=2$, these two curves intersect at $\theta=\frac{\pi}{12}$ and $\theta=\frac{5\pi}{12}$.

The formula you want to use is $\frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta$

Plugging in gives you $4\cdot\frac{1}{2}\int_{\pi/12}^{5\pi/12}(4\sin(2\theta)-2)\,d\theta$
Spoiler:
$= 2\bigg[-2\cos(2\theta)-2\theta\bigg]_{\pi/12}^{5\pi/12}$ $= 2\left[\left(\sqrt{3}-\frac{5\pi}{6}\right)-\left(-\sqrt{3}-\frac{\pi}{6}\right)\right] = 2\left(2\sqrt{3}-\frac{2\pi}{3}\right) = 4\sqrt{3}-\frac{4\pi}{3}$
Why
$\frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta$
but not
$\frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1 - r_2)\,d\theta$
?

5. That's just the formula for polar curves. It comes from double integrals.

$\int\int\,dA = \int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2( \theta)}r\,dr\,d\theta = \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_2^2-r_1^2)\,d\theta$

6. i guess i graphed this out wrong, so i'm not seeing why i have to calculate four areas and not just two. isn't the first curve a lemniscate with a 45 degree tilt, and the second one just a circle, making it only two areas i have to calculate for the area?

7. Originally Posted by tomobson
i guess i graphed this out wrong, so i'm not seeing why i have to calculate four areas and not just two. isn't the first curve a lemniscate with a 45 degree tilt, and the second one just a circle, making it only two areas i have to calculate for the area?
Yes, you're right. When I graphed the equation of the first curve, I forgot it was $r^2=4\sin(2\theta)$. (I graphed $r=4\sin(2\theta).)$ However, most of my calculations are still correct, I believe, except that instead of multiplying by 4 at the end, I should have multiplied by 2. I have edited my original post.

8. Originally Posted by redsoxfan325
That's just the formula for polar curves. It comes from double integrals.

$\int\int\,dA = \int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2( \theta)}r\,dr\,d\theta = \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_2^2-r_1^2)\,d\theta$

I see.
I haven't learn it yet.