\(\int\limits_{0}^{2a}\;\int\limits^{\sqrt{2ax}}_{\sqrt{2ax-x^2}}\;dx\,dy\)
We observed that strips in required region varies from \(y=\sqrt{2ax-x^2}\) to \(y=\sqrt{2ax}\) and region is bounded between x=0 to x=2a.
Now, \(y=\sqrt{2ax-x^2}\)
⇒ y2 = 2ax-x2 (squaring both sides)
⇒ x2-2ax+y2 = 0
⇒ x2-2ax+a2+y2 = a2 (By adding a2 in both sides)
⇒ (x-a)2+y2 = a2 ...(i)
Curve (i) is a circle whose centre is (a, 0) and radius a.
Now, \(y=\sqrt{2ax}\)
⇒ y2 = 2ax ....(2) (By squaring both sides)
Curve (2) is a parabola whose vertex is (0, 0) and axis of parabola is x-axis.
After drawing curve (i) and (2), we observed that shaded region for given integration divided into 3 regions when we took strips parallel to x-axis.
At curve (1) the abscissa (x-a)2 = a2-y2 (From (i))
⇒ x-a \(=\pm\sqrt{a^2-y^2}\)
⇒ x \(=a\pm\sqrt{a^2-y^2}\)
At curve (2) the abscissa x = y2/2a (from (2)).
Region I :- In region (1) the strip (Parallel to x-axis) varies from x = y2/2a (Parabola) to x \(=a-\sqrt{a^2-y^2}\)
(First half part of circle) and to cover whole first region, the strip must varies from y=0 to y=a,
Region II :- In region (2) the strip varies from x \(=a+\sqrt{a^2-y^2}\) (second part of circle) to x = 2a (line) and to cover the whole II region the strip must varies from y=0 to y=a.
Region III :- In region III, the strip varies from x = y2/2a (Parabola) to line x = 2a and to cover whole region III, the strip must be varies from y=a to y=2a.
∴ While changing in order of integration the integral change to
\(\int\limits_{0}^{2a}\;\int\limits^{\sqrt{2ax}}_{\sqrt{2ax-x^2}}\;dx\,dy\) \(=\int\limits_{0}^{a}\;\int\limits^{a-\sqrt{a^2-y^2}}_{\sqrt{x\,=\,y^2/2a}}\;dy\,dx\) \(+\int\limits_{0}^{a}\;\int\limits^{2a}_{a+\sqrt{a^2-y^2}}\;dy\,dx\) \(+\int\limits_{a}^{2a}\;\int\limits^{2a}_{y^2/2a}\;dy\,dx\)
