Question

Polar coordinates: Find the area of the region enclosed by one loop of the curve.

r=1+2sintheta(inner loop)

need help finding the limits of integration.

I get sintheta=-1/2, now where do i go from there?

Answer 1

@Hoa still not entirely sure. :l

Answer 2

ok

Answer 3

i figured it out actually. since sin theta=-1/2, i have to use the quadrants 3 and 4. i was using 1 and 2.

Answer 4

Not entirely sure what you mean to be honest with you, haha. :P

Answer 5

@mathslover @modphysnoob @Mathhater94 HELP in This question

Answer 6

I got it figured out, thanks for the help. going to close the question.

Answer 7

limits of integration were 7pi/6 and 11pi/6

Answer 8

I think it would be \(\cfrac{7 \pi}{6} \) to \(\cfrac{3\pi}{2}\)

Answer 9

3pi/2 is right on the y axis, would that still work?

Answer 10

The curve reaches the highest point of the inner loop at \(\cfrac{3\pi}{2}\) .

I had used symmetry here : the function can be evaluated from \(\cfrac{7\pi}{6}\) to \(\cfrac{3\pi}{2} \) and multiply by 2 to get the area of the entire loop.

Answer 11

\(2\int _ {\cfrac{7\pi}{6}} ^{\cfrac{3\pi}{2}} \cfrac{1}{2} (1+2\sin \theta)^2 d\theta\)
\(\implies \int _ {\cfrac{7\pi}{6}} ^{\cfrac{3\pi}{2}} 1 + 4\sin \theta +4(\cfrac{1-\cos 2\theta}{2})d\theta\)
\(\implies [3\theta +4\cos \theta -sin 2\theta ]_{\cfrac{7\pi}{6}}^{\cfrac{3\pi}{2}}\)
\(\textbf{which gives us} \space = \pi - \cfrac{3\sqrt{3}}{2}\)

Answer 12

If I am right then ^ above will be the correct work.