Question

Area of the circle x2+y2=16 is

Answer 1

$\large\underline{\sf{Solution-}}$

Method : - 1

Given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 16$

It represents the equation of circle whose center is ( 0, 0 ) and radius 4 units.

So, required area using integration is given by

$\rm :\longmapsto\:Area =4 \displaystyle\int_0^4\rm y \: dx$

$\rm \: = \:4\displaystyle\int_0^4\rm \sqrt{ {4}^{2} - {x}^{2} } \: dx$

We know,

$\boxed{ \bf{ \: \displaystyle\int\rm \sqrt{ {a}^{2} - {x}^{2} } \: dx = \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a}}}$

So, using this result, we get

$\rm \: = \:4\bigg( \dfrac{x}{2} \sqrt{ {4}^{2} - {x}^{2} } + \dfrac{ {4}^{2} }{2} {sin}^{ - 1} \frac{x}{4}\bigg)_0^4$

$\rm \: = \:4\bigg( \dfrac{4}{2} \sqrt{ {4}^{2} - {4}^{2} } + \dfrac{ {4}^{2} }{2} {sin}^{ - 1} \dfrac{4}{4} - 0 -0 \bigg)$

$\rm \: = \:4\bigg(0+ 8 {sin}^{ - 1} (1) \bigg)$

$\rm \: = \:4 \times 8 \times \dfrac{\pi}{2}$

$\rm \: = \:16\pi \: square \: units$

Alter Method :-

Given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 16$

We know,

Equation of circle having center ( h, k ) and radius r is given by

$\rm :\longmapsto\: \boxed{ \bf{ \: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} }}$

So, on comparing with this equation, we get r = 4 units.

Now,

$\rm :\longmapsto\:Area_{(circle)} = \pi {r}^{2}$

$\rm :\longmapsto\:Area_{(circle)} = \pi {(4)}^{2}$

$\bf :\longmapsto\:Area_{(circle)} = 16 \: \pi \: square \: units$