Using integration, find the area of the region : {(x,y):0 ≤ 2 ≤ , 0 ≤ ≤ , 0 ≤ ≤ 3}
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Answer:
x
2
+y
2
−2ax+a
2
=a
2
is a circle, which can also be written as (x−a)
2
+y
2
=a
2
Center : (a,0) and radius = a
The region given is that inside the circle.
y
2
=ax is a parabola with its face opening towards positive infinity.
y
2
≥ax is the area outside the parabola.
The common area is thus inside the circle, outside the parabola and in the first quadrant.
The intersection point is obtained by x
2
+ax−2ax+a
2
=a
2
i.e. x=a,0
Area =∫
0
a
[x
2
+y
2
−2ax−y
2
+ax]dx
Area =∫
0
a
[x
2
−2ax+ax]dx
=
0
a
[
3
x
3
−
2
ax
2
]
=
3
a
3
−
2
a
3
Taking its modulus, Area=
6
a
3
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