Calculus II

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First we need to calculate intersection points. This we can determine from these two equations

`y=-t-3/t`

`y=4`

`-t-3/t=4`

`(-t^2-3)/t=4`

`t^2+4t+3=0`

`t_1,2=(-4pmsqrt(16-4cdot1cdot3))/2`

`t_1=-3`

`t_2=-1`

Also since `x(t_1)=-2` and `x(t_2)=2` it's easy to see that area under `y=4` (this is simple to calculate because it is actually square) is `4cdot(2-(-2))=16.`  If we subtract area under the other curve  we will get area between the curves.

And since area under parametrically defined curve `x=f(t),` `y=g(t)`  is given by formula

`A=int_(t_1)^(t_2)g(t)f'(t)dt`  

we have

`A=16-int_-3^-1(-t-3/t)(1+3/t^2)dt=`

`16-int_-3^-1(-t-3/t-3/t-9/t^3)dt=`

`16+(t^2-6ln|t|-9/(2t^2))|_-3^-1=`

`16+1/2+6cdot0-9/2-9/2-6ln3+1/2=`

` ` `8-6ln3` <--Your solution

Area between the curves is equal to `8-6ln3.`

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