Area between the curves

Question

Find the area between !!x^2=y, x^2=-y and y^2=4x-3!!

!!a" "4/3!!

!!b" "5/3!!

!!c" "2/3!!

!!d" "1/3!!

Solution

In order to find the area between the curves we need to first draw its !!2D!! coordinate diagram.

We can find area between the curves using both !!x!! axis and !!y!! axis

Method !!1!!

If we will find the area between the curves using !!x!! axis.

We can see that shaded area above !!x!! is the area of upper curve minus lower curve.

Hence, we need to first get the point of intersection between curves !!x^2=y and y^2=4x-3!! will be !!(1,1)!!

Similarly, point of intersection between curves !!x^2=-y and y^2=4x-3!! will be !!(1,-1)!!.

Also vertex point of curve !!y^2=4x-3!! is !!(3/4,0)!!

Since curves above and below the !!x!! axis are same.

Hence we can find the area between the curves !!x^2=y and y^2=4x-3!! above the !!x!! axis and by multiplying it with !!2!! we will get required area.

Area = !!2[ int_0\^1x^2dx - int_(3/4)\^1sqrt(4x-3)dx]!!

!!= 2[ (x^3/3)_0\^1 - ((1/4)((4x-3)^(3/2))/(3/2))_(3/4)\^1]!!

!!= 2[ (1/3 - 0) - (1/6)(4x-3)_(3/4)\^1]!!

!!= 2[ 1/3 - (1/6)(1-0)]!!

!!= 2[1/3 - 1/6]

!!= 2[1/6]!!

!!= 1/3!!

Method !!2!!

On the !!y!! axis we can substract the area of the curve !!y=x^2!! from !!y^2=4x-3!!

Area = !!2[ int_0\^1(y^2+3)/4dy - int_0\^1sqrtydy]!!

!!= 2[ (1/4)((y^3/3)+3y)_0\^1 - (2/3)(y^(3/2))_0\^1]!!

!!= 2[ (1/4)(1/3 + 3) - (2/3)(1-0)]!!

!!= (1/2)(10/3) - (4/3)!!

!!= 5/3 - 4/3!!

!!= 1/3!!

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