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Find the area of region in xy-plane, defined by inequalities x-2y^2>equal 0,...

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elahusb | Student, College Freshman | (Level 1) eNoter

Posted May 7, 2012 at 2:53 PM via web

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Find the area of region in xy-plane, defined by inequalities x-2y^2>equal 0, 1-x-IyI>equal 0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 7, 2012 at 4:22 PM (Answer #1)

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You need to solve the inequalitites such that:

`x - 2y^2gt= 0 =gt x gt= 2y^2`

`1-x-|y| gt= 0 =gt -x gt= |y| - 1=gt x =lt 1 - |y|`

Hence, the region is bounded by 3 curves `x = 1+y, x=1-y, x=2y^2` .

You need to determine the points of intersection between curves `x = 1+y`  and `x=2y^2`  such that:

`1 + y = 2y^2 =gt 2y^2 - y - 1 = 0`

You need to solve quadratic equation such that:

`y_(1,2) = (1+-sqrt(1 + 8))/4`

`y_(1,2) = (1+-3)/4 =gt y_1 = 1`

`y_2 = -1/2`

You need to determine the points of intersection between curves `x= 1-y`  and `x=2y^2`  such that:

`1 - y = 2y^2 =gt 2y^2+ y - 1 = 0`

You need to solve quadratic equation such that:

`y_(1,2) = (-1+-sqrt(1 + 8))/4`

`y_(1,2) = (-1+-3)/4 =gt y_1 = -1`

`y_2 = 1/2`

Hence, the limits of integration are `-1/2`  and `1/2`  and you need to evaluate the area of the region modelled by function `1 - y - 2y^2`  such that:

`int_(-1/2)^(1/2) (1 - y - 2y^2 ) dy`

Since the limits of integration ar opposed numbers, hence you need to evaluate the definite integral such that:

`2int_0^(1/2) (1 - y - 2y^2 ) dy = 2(y - y^2/2 - 2y^3/3)|_0^(1/2)`

`2int_0^(1/2) (1 - y - 2y^2 ) dy = 2(1/2 -1/8- 1/12)`

`2int_0^(1/2) (1 - y - 2y^2 ) dy = (12 - 3- 2)/12`

`2int_0^(1/2) (1 - y - 2y^2 ) dy = 7/12`

Hence, evaluating the area of the region under given conditions yields`int_(-1/2)^(1/2) (1 - y - 2y^2 ) dy = 7/12.`

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