Question

An ellipse is cutout of a circle of radius a , the major axis of the ellipse coincides with one of the diameter of the circle while the minor axis is equal to 2b . Prove that the area of the remaining art equals that of the ellipse with the semi axes a and a-b

Answer 1

we have,

equation of circle is

$x^2+y^2=a^2......\left(1\right)$

$A_{circle}=4{\int }_0^{a}x.dy$

Also, from eqn (1)

$x=\sqrt{a^2-y^2}$

So,

$A_{circle}=4{\int }_0^a\sqrt{a^2-y^2}dy$

Now, put

$y=a\sin \theta$

Then,

$\frac{dy}{d\theta }=a\cos \theta$

$dy=a\cos \theta d\theta$

Change limit,

Also, as

$y=0,\theta =0$

As $y=a$$\theta =\frac{\pi }{2}$

Further,

$A_{circle}=4{\int }_0^{\frac{\pi }{2}}\sqrt{a^2-a^2{\sin }^2\theta }\cos \theta d\theta$

$A_{circle}=4a{\int }_0^{\frac{\pi }{2}}\sqrt{a^2(1-{\sin }^2\theta )}\cos \theta d\theta$

$A_{circle}=4a^2{\int }_0^{\frac{\pi }{2}}\sqrt{{\cos }^2\theta }\cos \theta d\theta$

$A_{circle}=4a^2{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

Also we know that,

$\cos 2\theta =2{\cos }^2\theta -1$

${\cos }^2\theta =\frac{\cos 2\theta +1}{2}$

Therefore,

$A_{circle}=4a^2{\int }_0^{\frac{\pi }{2}}\left(\frac{\cos 2\theta +1}{2}\right)d\theta$

$A_{circle}=2a^2{\int }_0^{\frac{\pi }{2}}\cos 2\theta d\theta +{\int }_0^{\frac{\pi }{2}}1d\theta$

$A_{circle}=2a^2{{(\frac{\sin 2\theta }{2}+\theta )}_0}^{\frac{\pi }{2}}$

$A_{circle}=2a^2(\frac{\sin \pi -\sin 0}{2}+\frac{\pi -0}{2})$

$A_{circle}=\pi a^2$

Hence, this is the answer.