Find the area of the region included between the parabolas `y^2=4a xa n dx^2=4a y ,w h e r ea > 0.`

Explanation

For finding the points of intersection of the two parabolas, we equate the values of ` y^2 ` from their equations.
From the equation
` x ^2 = 4 a y , y = ( x ^2) / ( 4a ) `
` therefore y ^2 = ( x ^ 4 ) / ( 16 a ^2) = 4ax `
` therefore x ^4 - 64 a ^ 3 x = 0 " "x ( x ^3 - 64 a ^ 3 ) = 0 `
` therefore x = 0 , x = 4a `

When ` x = 0, y = 0 `
When ` x = 4a, y = 4a`
` therefore ` the points of intersection are ` O ( 0, 0) and A ( 4a, 4a )`.
Required area = area of the region OBACO
= ( area of the region ODACO ) - ( area of the region ODABO)
Now, area of the region ODACO = area under the parabola ` y ^(2) = 4 ax `, i.e., ` y = 2 sqrta * sqrt x `
` = int _ 0 ^ ( 4a ) 2 sqrta * sqrt x dx = 2 sqrt a int _ 0 ^ ( 4a) x ^((1)/(2)) dx `
` = 2 sqrt a [ ( x^((3)/(2))) /( 3 //2)] _ 0 ^ 4 a = ( 4 sqrt a ) /(3) = [ ( 4a ) ^((3) /(2)) - 0 ] = ( 32 a ^(2))/( 3 ) `
Area of the region ODABO = area under the parabola ` x ^2 = 4 ay `, i.e., ` y = ( x ^2)/( 4a )`
` = int _ 0 ^ (4a) ( x ^(2))/( 4a) dx = ( 1) /( 4a) int _ 0 ^( 4a ) x ^2 dx `
` = ( 1 ) /( 4a ) [ ( x^(3))/( 3) ] _ 0 ^ ( 4a ) = ( 1 ) / ( 12 a ) [ ( 4a) ^(3) - 0 ] `
` = ( 1 ) / ( 12a ) xx 64 a ^ 3 = ( 16 a ^2)/( 3 ) `
` therefore ` required area = ` ( 32a ^(2))/(3) - ( 16 a ^ 2 )/ ( 3 ) = ( 16 a ^2) / ( 3 ) ` sq units.