Draw a rough sketch of the region and find the area enclosed by the region, using the method of integration.


Given the boundaries of the area to be found are,


R = {(x,y): y2 ≤ 3x, 3x2 + 3y2 ≤ 16}


This can be written as


R1 = {(x,y): y2 ≤ 3x}


R2 = {(x,y): 3x2 + 3y2 ≤ 16}


Then


From R1, we can say that , y2 = 3x is a parabola


y2 = 3x ---- (1)


• With vertex at (0,0) i.e. the origin


• Symmetric about the x-axis, as it has the even power of y


From R1, we can say that , 3x2 + 3y2 = 16 is a circle


3x2 + 3y2 = 16 ----- (2)


• the vertex at (0,0) i.e. the origin


• the radius of units


Now to find the point of intersection of (1) and (2), substitute y2 = 3x in (2)


3x2 + 3(3x) = 16


3x2 + 9x - 16= 0



as y cannot be imaginary, we reject the negative value of x


so


So the two points, A and B are the points where (1) and (2) meet.


The line AB meets the x-axis at


Substitute y = 0 in (2),


3x2 + 0 = 16



So the circle intersects the x-axis at and



As x and y have even powers for the circle, they will be symmetrical about the x-axis and y-axis.


Consider the circle, 3x2 + 3y2 = 16, can be re-written as



----- (3)


Consider the parabola, y2 = 3x, can be re-written as


----- (4)


Now, the area to be found will be the area is


Area of the required region = Area of OACBO.


Area of OABCO= Area of OCAO+ Area of OCBO


[area of OCBO= area of OCAOas the circle is symmetrical about the y-axis]


Area of OACBO= 2 × Area of OCAO---- (5)


Area of OCAO= Area of OADO+ Area of DACD


Area of OCAOis




[Using the formula, and ]


Let




[sin-1(1) = 90° and sin-1(0) = 0°]




The Area of OCAOsq. units, where


Now substituting the area of OCAOin equation (5)


Area of OACBO= 2 × Area of OCAO




Area of the required region is sq. units, where


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