Area Inside Polar curves

[shivers20]

Inside the six-leaved rose r^2 = 2 sin 3 theta

Sketch the area:

Make a table to get points. Done!

d A= 1/2 r^2 d theta

Integrate:

1/2 (2sin 3 theta)^2 d theta

Area= (0-2pi) 2 | (sin 3 theta)^2 d theta

use symmetry:

A= 2 | (0-pi) 2(sin 3 theta)^2 d theta.

Am I on the right track? Can I end it here?

 

[skeeter]

why did you square r² in the integrand? your polar equation is
\(\displaystyle \L r^2 = 2\sin{3\theta}\)

\(\displaystyle \L \frac{1}{2}r^2 d\theta = \sin{3\theta} d\theta\)

you have six "petals" ... find the area of one and multiply by six ...

\(\displaystyle \L A = 6\int_0^{\frac{\pi}{3}} \sin{3\theta}d\theta\)

 

[shivers20]

why did you square r² in the integrand? your polar equation is
\(\displaystyle \L r^2 = 2\sin{3\theta}\)

\(\displaystyle \L \frac{1}{2}r^2 d\theta = \sin{3\theta} d\theta\)

you have six "petals" ... find the area of one and multiply by six ...

\(\displaystyle \L A = 6\int_0^{\frac{\pi}{3}} \sin{3\theta}d\theta\)

I see where I made the mistake of squaring it when r^2 = 2sin3 theta.
Heres a dumb question, how did you get |0 - pi/3?

 

[skeeter]

the graph starts at the origin at theta = 0

one "petal" starts at the origin and goes back to it ... how much angle do you need to get back to the origin if the argument for the sine function is 3*theta?

 

[shivers20]

the graph starts at the origin at theta = 0

one "petal" starts at the origin and goes back to it ... how much angle do you need to get back to the origin if the argument for the sine function is 3*theta?

I have this example in my notes, r= 2 sin 2 theta
2 sin 2theta= 1
sin 2 theta = 1/2
2theta= pi/6
theta= pi/12
I tried following the same steps but it gave me the wrong answer

 

[soroban]

Hello, shivers20!

Area formula: \(\displaystyle \L\:A\;=\;\frac{1}{2}\int^{\;\;\;\beta}_{\alpha}r^2\,d\theta\)

Inside the six-leaved rose: \(\displaystyle \,r^2 \:= \:2\,\sin3\theta\)

The graph is a six-leaf rose.
The six equal leaves are equally spaced (60° apart).

When \(\displaystyle x\,=\,0:\;r^2\,=\,0\)
When \(\displaystyle x\,=\,\frac{\pi}{6}:\;r^2\,=\,2\)
When \(\displaystyle x\,=\,\frac{\pi}{3}:\:r^2\,=\,0\)

Hence, one leaf is generated from \(\displaystyle \theta\,=\,0\) to \(\displaystyle \frac{\pi}{3}\)

The area of one leaf is: \(\displaystyle \L\:\frac{1}{2}\int^{\;\;\;\frac{\pi}{3}}_02\,\sin3\theta\,d\theta\)

Evaluate . . . and multiply by 6.

 

[skeeter]

I have this example in my notes, r= 2 sin 2 theta
2 sin 2theta= 1
sin 2 theta = 1/2
2theta= pi/6
theta= pi/12
I tried following the same steps but it gave me the wrong answer

this example is a 4-petal rose ... the graph starts at the origin and returns to the origin after pi/2 radians. in other words, one petal on the graph is completed after theta runs from 0 to pi/2 radians.

area of one petal ...

\(\displaystyle \L A = \int_0^{\frac{\pi}{2}} \frac{1}{2} 4\sin^2{2\theta} d\theta\)

multiply that by 4 to get the total area ...

\(\displaystyle \L A = 8\int_0^{\frac{\pi}{2}} \sin^2{2\theta} d\theta\)

 

[galactus]

\(\displaystyle r^{2}=2sin(3{\theta})\)

In other words we must graph:

\(\displaystyle r=\sqrt{2sin(3{\theta})};\ and \;\ r=-\sqrt{2sin(3{\theta})}\)


\(\displaystyle \L\\3\int_{0}^{\frac{\pi}{3}}(2sin(3{\theta}))d{\theta}=4\)

EDIT: I redone the graph after the cerebral flatulence.

 

[skeeter]

May I ask why we would multiply by 6 and not 3?.

have you graphed \(\displaystyle \L r^2 = 2\sin{3\theta}\)?

 

[galactus]

That's the graph I just posted.

 

[soroban]

Hello, Cody!

You graphed \(\displaystyle r\:=\:2\sin3\theta\), a three-leaf rose curve,

. . .not \(\displaystyle \,r^2\:=\:2\sin3\theta\), a six-leaf rose curve.
. . . . . . .↑

You are correct: \(\displaystyle \:r\:=\;a\cdot\sin(n\theta)\text{ has: }\begin{Bmatrix}n\text{ leaves } & \text{ odd }n \\ 2n\text{ leaves } & \text{ even }n\end{Bmatrix}\)

So we can have rose curves with: 1, 3, 5, 7, ..., (2n-1) leaves
. . . . . . . . . . . . . . . . . . . . and: 4, 8, 12, 16, ..., 4n leaves.


But to obtain the other even-leafed rose curves (2, 6, 10, 14, ... leaves),

. . we use: \(\displaystyle \:r^2\:=\:a\cdot\sin(n\theta)\;\) for odd \(\displaystyle n.\)

 

[galactus]

DUH, DUH, DUH. I just seen my idiocy as you posted, Soroban. I see.

Brain fart . Thanks for the reply. I overlooked the r^2, wasn't thinkin'.