Find the area of the region. r2=sin⁡(2θ)

Question

Find the area of the region.    r2=sin⁡(2θ)

habbocowji · Accepted Answer

Firstly, note that the graph of the Lemniscate

r^{2} = \sin (2 θ)

is a symmetric with respect to the origin as follows
Since

(r, θ)

on the graph

\Rightarrow (- r, θ)

is on the graph, thus

(\pm r^{2}) = \sin (2 θ) \Rightarrow r^{2} = \sin (2 θ)

And the garph is not symmetric about the x-axis, therefore the graph is not symmetric about y-axis for the following reasons:

\sin 2 (- θ) = - \sin (2 θ) \neq r^{2}

\sin 2 (π - θ) = \sin (2 π - 2 θ))

= \sin (- 2 θ)

= - \sin (2 θ)

\neq r^{2}

Thus, the graph must be symmetric with respect to the origin.
The area of the region between theorign and the curve

r = f (θ)

, where

α \leq θ \leq β

is

A = \int_{α}^{β} \frac{1}{2} r^{2} d θ

We can see that the lemniscate is symmetric about the origin and the loop of the curve lines between

θ = 0

to

\frac{π}{2}

so that the area of the region is

A = 2 \int_{0}^{\frac{π}{2}} \frac{1}{2} r^{2} d θ

= \int_{0}^{\frac{π}{2}} r^{2} d θ

Substitute now by the polar equation of the lemniscate

r^{2} = \sin (2 θ)

. Therefore< the previous equation becomes:

A = \int_{0}^{\frac{π}{2}} \sin (2 θ) d θ

= {[- \frac{\cos (2 θ)}{2}]}_{0}^{\frac{π}{2}}

= - \frac{1}{2} {[\cos (2 θ)]}_{0}^{\frac{π}{2}}

= - \frac{1}{2} [\cos 2 (\frac{π}{2}) - \cos 2 (0)]

= - \frac{1}{2} [\cos (π) - \cos (0)]

= - \frac{1}{2} [- 1 - 1]

= - \frac{1}{2} [- 2]

= 1

Note that the area inside the one loop of the LEmniscate

r^{2} = \sin 2 θ

which represents the shaded area of the graph is given by:

A = \int_{0}^{\frac{π}{2}}

aquariump9 · Accepted Answer

The total area of region of the Lemniscate r2=sin⁡(2θ) is A=2∫0π212r2dθ=1

Find the area of the region. r^{2}=\sin(2\theta)