How do you find the area of the region bounded by the polar curves $r^{2} = cos (2 θ)$ $r^2=cos(2theta)$ and $r^{2} = sin (2 θ)$ $r^2=sin(2theta)$ ?

Question

10912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-11-02T12:59:43

Let us look at the region.

enter image source here

(Note: $r^{2} = cos 2 θ$ $r^2=cos2theta$ in purple, and $r^{2} = sin 2 θ$ $r^2=sin2theta$ in blue)

Since there two identical region, we will find a half of one region then multiply by $4$ $4$ . The combine area $A$ $A$ can be found by

$A = 4 \int_{0}^{\frac{π}{8}} \int_{0}^{\sqrt{2 sin 2 θ}} r d r d θ$ $A=4int_0^{pi/8}int_0^{sqrt{2sin2theta}}rdrd theta$

$= 4 \int_{0}^{\frac{π}{8}} {[\frac{r^{2}}{2}]}_{0}^{\sqrt{sin 2 θ}} d θ$ $=4int_0^{pi/8}[r^2/2]_0^{sqrt{sin2theta}}d theta$

$= 2 \int_{0}^{\frac{π}{8}} sin 2 θ d θ$ $=2 int_0^{pi/8}sin2theta d theta$

$= 2 {[\frac{- cos 2 θ}{2}]}_{0}^{\frac{π}{8}}$ $=2[{-cos2theta}/2]_0^{pi/8}$

$= - cos (\frac{π}{4}) + cos (0)$ $=-cos(pi/4)+cos(0)$

$= - \frac{\sqrt{2}}{2} + 1 = \frac{2 - \sqrt{2}}{2}$ $=-{sqrt{2}}/{2}+1={2-sqrt{2}}/2$

I hope that this was helpful.

How do you find the area of the region bounded by the polar curves r^2=cos(2theta)r2=cos(2θ)r^2=cos(2theta) and r^2=sin(2theta)r2=sin(2θ)r^2=sin(2theta) ?