Question

Find the area inside the oval limaçon r = 6+sin θ

Answer 1

as i recall it is
\[\frac{1}{2}\int r^2 d\theta\]

Answer 2

limits of integration from \(0\) to \(2\pi\)

Answer 3

Hm, I'll try that. I seem to be using the wrong limits.

Answer 4

since this is \(\sin(\theta)\) and not say \(\sin(2\theta)\) or some such thing, once complete rotation is from 0 to \(2\pi\)

Answer 5

i think it is \(\frac{73\pi}{2}\) but i wouldn't bet money on it

Answer 6

That's actually the correct answer - I've been sort of trying to backwards engineer it. I keep ending up with the wrong answer. I think I"m integrating wrong somehow.
I keep ending up with \[37/2\theta - 6\cos \theta+(\sin2\theta)/4\]

Answer 7

It looks like I'm dropping a 1/2 somewhere along the line - I think the formula should be \[73\theta/4 -6\cos \theta-1/8\sin2\theta \]

Answer 8

\[\sin^2(\theta)=\frac{1}{2}(1-\cos(2\theta)\] so
\[\int\sin^2(\theta)=\frac{\theta}{2}-\frac{1}{4}\sin(2\theta)\]

Answer 9

\[36+12\sin(\theta)+\sin^2(\theta)\]
anti derivative is
\[36\theta-12\cos(\theta)+\frac{\theta}{2}-\frac{1}{4}\sin(2\theta)\]

Answer 10

\[1/2\int\limits_{0}^{2\pi}(6+\sin)^2\]
\[=1/2\int\limits_{0}^{2\pi}36+12\sin \theta+\sin^2\]
\[= 1/2(36\theta - 12\cos \theta +\int\limits \sin^2\theta)\]
\[=1/2 (36\theta - 12\cos \theta+\int\limits (1-\cos2\theta)/2\]
\[= 1/2(36\theta-12\cos \theta+\theta/2-(\sin2\theta))/2\]
\[=37/2\theta-6\cos \theta-(\sin2\theta)/4\]

Answer 11

when you plug in 0 you get -12
when you evaluate at \(2\pi\) you get
\[72\pi+\pi-12\]

Answer 12

subtract and get \(73\pi\) then divide by 2

Answer 13

oh my god, I have been forgetting to multiply the 36 by 2 before adding it to the 1/2.

Answer 14

Thank you so much for all your help - I've been working on this one problem for hours.