Given equations are :
x = 1 (represents a line parallel to y - axis at a distance 1 to the right)
equation (1) represents a half eclipse that is symmetrical about the x - axis and also about the y - axis with center at origin.
A rough sketch is given as below: -
We have to find the area of shaded region.
Required area
= (shaded region OABCO)
(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)
\(=\int^1_0 y\,dx\) (As x is between (0,1) and the value of y varies)
Substitute x = a sin u \(\Rightarrow\) u = sin-1(\(\frac{x}{a}\)), dx = a cos u du
So the above equation becomes,
So the above equation becomes,
Apply reduction formula:
Undo the substituting, we get
On applying the limits we get,
Hence the area under the y = \(\sqrt{a^2-x^2}\) included between the lines x = 0 and x = 1 is equal to \(\frac{x^2}{2}[(\text{sin}^{-1}(\frac{1}{a})+\frac{1}{a^2}\sqrt{(a^2-1)}]\) square units.
