# Please help???

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### 1 Answer

The right hand loop graph of the lemniscate is attached below. The domain of angle variation is

`-pi/2 <=theta<pi/2`

The corresponding function is

`r =sqrt(cos(2theta))`

Thus

`r =cos(2theta)` and `2rdr =-2sin(2theta)d(theta) rArr (dr)/(d theta) =-sin(2theta)/r =-sin(2theta)/sqrt(cos(2theta))`

The surface area of the right hand loop is

`S =int_(-pi/2)^(pi/2) 2*pi*r*cos(theta)sqrt(cos(2theta)+(sin^2(2theta))/cos(2theta)) d(theta)`

`S =int_(-pi/2)^(pi/2) 2*pi*r*cos(theta)*1/sqrt(cos(2theta))d(theta) =2*pi*int_(-pi/2)^(pi/2)cos(theta)*d(theta) =`

`=2*pi*sin(theta) (-pi/2->pi/2) =4*pi`

**Answer: The surface area of the object of rotation is `S =4*pi` **