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