`4x + y^2 = 12, x = y` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

Expert Answers

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You need to determine first the points of intersection between curves `y = sqrt(12-4x)` and `y = x` , by solving the equation, such that:

`sqrt(12-4x)= x =>12-4x = x^2 => x^2 + 4x - 12 = 0`

`x_(1,2) = (-4+-sqrt(16 + 48))/2 => x_(1,2) = (-4+-8)/2`

`x_1 = 2; x_2 = -6`

Hence, the endpoints of integral are x = -6 and x = 2

You need to decide what curve is greater than the other on the interval [-6,2]. You need to notice that `sqrt(12-4x) >x` on the interval [-6,2],hence, you may evaluate the area of the region enclosed by the given curves, such that:

`int_a^b (f(x) - g(x))dx` , where f(x) > g(x) for `x in [a,b]`

`int_(-6)^2 (sqrt(12-4x) - x)dx =int_(-6)^2sqrt(12-4x) dx - int_(-6)^2 xdx`

You need to solve `int_(-6)^2sqrt(12-4x) dx` using substitution `12 - 4x = t` , such that:

`12-4x = t => -4dx = dt => dx = -(dt)/4`

`int_(-6)^2sqrt(12-4x) dx= -(1/4)int_(t_1)^(t_2)sqrt(t) dt`

`int_(-6)^2sqrt(12-4x) dx= -2/12 (12-4x)^(3/2)|_(-6)^2`

`int_(-6)^2sqrt(12-4x) dx= -2/12 (12-4*2)^(3/2) + 2/12(12 + 24)`

`int_(-6)^2sqrt(12-4x) dx= -(2/12)*8 + (36*2)/12`

`int_(-6)^2sqrt(12-4x) dx= - 4/3 + 6`

`int_(-6)^2sqrt(12-4x) dx= (18-4)/3`

`int_(-6)^2sqrt(12-4x) dx= 14/3`

Hence, evaluating the area of the region enclosed by the given curves, yields `int_(-6)^2sqrt(12-4x) dx= 14/3.`

The area of the region enclosed by the given curves is found between the red and orange curves, for `x in [-6,2].`

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