Find the exact area enclosed between the curve y=sqrt(4-x^2) and the line x-y+2=0

sciencesolve | Certified Educator

You first need to find the interception points between the curves, hence, you should solve the system of equations `y=x+2`  and `y=sqrt(4-x^2)`  such that:

`x+2 = sqrt(4-x^2)`

You need to raise to square to remove the square root such that:

`(x+2)^2 = 4 - x^2`

You need to expand the binomial such that:

`x^2 + 4x + 4 = 4 - x^2`

You need to move all terms to the left side such that:

`x^2 + 4x + 4 - 4 + x^2 = 0`

`2x^2 + 4x = 0`

Factoring out 2x yields:

`2x(x + 2) = 0 =gt x = 0 and x+2 = 0 =gt x = -2`

You need to verify what is the upper limit curve and what is the lower limit curve, between -2 and 0 values, hence, you should select the value -1 to evaluate both functions such that:

`y = -1+2 = 1`

`sqrt(4 - 1) = sqrt 3`

Notice that sqrt>1, hence the area is upper bounded by the curve `y=` `sqrt(4-x^2)`  and lower bounded by `y=x+2` , over the interval [-2,0].

You need to evaluate the definite integral of the function `sqrt(4-x^2) - x - 2 ` over [-2,0].

`int_(-2)^0 (sqrt(4-x^2) - x - 2) dx = int_(-2)^0 sqrt(4-x^2) dx - int_(-2)^0 x dx- int_(-2)^0 2 dx`

You need to substitute x by `2sint`  such that:

`x = 2sin t =gt dx = 2 cos t dt`

`sqrt(4-x^2) = (sqrt4(1-(4sin^2t)/4)`

`sqrt(4-x^2) = 2sqrt(1 - sin^2 t)`

`sqrt(4-x^2) = 2sqrt(cos^2 t)`

`sqrt(4-x^2) = 2 cos t`

Hence, evaluating the integral `int_(-2)^0 sqrt(4-x^2) dx`  using trigonometric substitution yields:

`int_(-2)^0 sqrt(4-x^2) dx = int^(-pi/2)_0 2 cos t*2 cos t dt`

int_(-pi/2)^0 2 cos t*2 cos t dt = 4int_(-pi/2)^0 cos^2 t dt

You need to use half angle formula such that:

`cos^2 t = (1 + cos 2t)/2`

`4int_(-pi/2)^0 cos^2 t dt = 4int_(-pi/2)^0(1 + cos 2t)/2 dt`

`2int^(-pi/2)_0 (1 + cos 2t) dt = 2int_(-pi/2)^0 dt + 2int_(-pi/2)^0 cos 2t dt `

`2int_(-pi/2)^0 (1 + cos 2t) dt = 2t|_(-pi/2)^0) + 2*(sin 2t)/2|_(-pi/2)^0`

`2int_(-pi/2)^0 (1 + cos 2t) dt = 2*pi/2 + sin 0 - sin(-pi/2)`

`2int_(-pi/2)^0 (1 + cos 2t) dt = pi + 0 + sin (pi/2)`

`2int_(-pi/2)^0 (1 + cos 2t) dt = pi + 1`

`int_(-2)^0 (sqrt(4-x^2) - x - 2) dx = pi + 1 - x^2/2|_(-2)^0 - 2x|_(-2)^0`

`int_(-2)^0 (sqrt(4-x^2) - x - 2) dx = pi + 1 - 0 + ((-2)^2)/2 - 0-4 `

`int_-2^0 (sqrt(4-x^2) - x - 2) dx = pi + 1 + 2 - 4`

`int_-2^0 (sqrt(4-x^2) - x - 2) dx = pi- 1`

Hence, evaluating the area enclosed between the curves yields `int_(-2)^0 (sqrt(4-x^2) - x - 2) dx = pi - 1.`