Question

How do you find the area of one petal of `r=6sin2theta`?

Answer 1

`(9pi)/2`

Explanation:

Area in polar coordinates is given by:

`A = int_alpha^beta 1/2 r^2 \ d theta`

The first step is to plot the polar curve to establish the appropriate range of `theta`

From the graph we can see that for the petal in Q1 then `theta in [0, pi/2]`

Hence,

`A = int_0^(pi/2) 1/2 (6sin2theta)^2 \ d theta`
`\ \ \ = 18 \ int_0^(pi/2) sin^2 2theta \ d theta`
`\ \ \ = 18 \ int_0^(pi/2) 1/2(1-cos 4 theta) \ d theta`
`\ \ \ = 9 \ int_0^(pi/2) 1-cos 4 theta \ d theta`
`\ \ \ = 9 \ [ theta -1/4sin 4 theta ]_0^(pi/2)`
`\ \ \ = 9 \ {(pi/2-1/4sin(2pi)) - (0)}`
`\ \ \ = (9pi)/2`

Answer 2

`9/2pi` areal units.

Explanation:

`r=sqrt(x^2+y^2)=6sin2theta >= 0`.

So, `2theta in Q_1 or Q_2`, giving `theta in Q_1`

The period of `sin 2theta` is `(2pi)/2=pi`

So, there are `(2pi)/pi=2` petals, for #theta in [0. 2pi]

The ares in one petal ( start at `theta = 0` and end at `theta =pi/2`

`=1/2 int r^2 d theta`, with`r = 6sin2theta and theta` from `0 to pi/2`

`=18 int sin^2(2theta) d theta`, for the limits

`=9 int (1-cos4theta) d theta`, for the limits

`=9[theta-1/4sin4theta]`, between 0 and `pi/2`.

`=9/2pi`

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