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!!