Find the area that lies inside both curves r = a and r2 = 2a2 cos(2θ). Here a is a positive constant.

### 1 Answer | Add Yours

Curves are `r=a` and `r^2=2a^2cos(2theta).`

Thus

`a^2=2a^2cos(2theta)`

`cos(2theta)=1/2`

`=> theta=pi/6,(5pi)/6`

Thus required area A,

`A=int_(pi/6)^((5pi)/6)(1/2)(2a^2cos(2theta)-a^2)d theta`

`` `=(a^2/2)(sin(2theta)-theta)_(pi/6)^((5pi)/6)`

`=(a^2/2){-sqrt(3)-(2pi)/3}`

Since area is not considered as negative, ignore negative sign.

Thus

`A=(sqrt(3)+(2pi)/3)(a^2/2)`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes