Question

Find the area of the smaller segment cut off from the area bounded by the circle x²+y²=16 by x+y =4​

Answer 1

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

Given curves are

$\rm \: {x}^{2} + {y}^{2} = 4 - - - (1)$

and

$\rm \: x + y = 4 - - - (2)$

Step :- 1 Point of intersection :

From equation (2) we have y = 4 - x

Substitute this value of y in equation (1), we get

$\rm \: {x}^{2} + {(4 - x)}^{2} = 16$

$\rm \: {x}^{2} + {x}^{2} + 16 - 8x = 16$

$\rm \: 2 {x}^{2} - 8x = 0$

$\rm \: 2x(x - 4) = 0$

$\rm\implies \:x = 0 \: \: or \: \: x = 4$

Hᴇɴᴄᴇ,

➢ Pair of points of intersection of equation (1) and (2) are shown in the below table :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

Step :- 2 Curve Sketching

Equation (1) represents the equation of circle having centre (0, 0) and radius 4 units.

Equation (2) represents the equation of line which passes through the points (0, 4) and (4, 0)

See the attachment.

Step :- 3 Evaluation of Area

So, required smaller area bounded between two curves with respect to x axis is

$\rm \: = \displaystyle\int_{0}^{4}\rm\ [y_{(circle)} - y_{(line)}] \: dx$

$\rm \: = \displaystyle\int_{0}^{4}\rm\ [ \sqrt{16 - {x}^{2} } - (4 - x )] \: dx$

$\rm \: = \displaystyle\int_{0}^{4}\rm\ [ \sqrt{ {4}^{2} - {x}^{2} } - (4 - x )] \: dx$

$\rm \: = \bigg[\dfrac{x}{2} \sqrt{ {4}^{2} - {x}^{2} } + \dfrac{ {4}^{2} }{2} {sin}^{ - 1} \dfrac{x}{4} - 4x + \dfrac{ {x}^{2} }{2} \bigg]_{0}^{4}$

$\rm \: = \bigg[0 + \dfrac{ {4}^{2} }{2} {sin}^{ - 1} \dfrac{4}{4} - 16 + \dfrac{ 16 }{2}\bigg] - (0 + 0 - 0 + 0)$

$\rm \: = 8 \times \dfrac{\pi}{2} - 16 + 8$

$\rm \: = 4\pi - 8$

$\rm \: = 4(\pi - 2) \: square \: units$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Required \: area \: \rm \: = 4(\pi - 2) \: square \: units \: \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

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

$\boxed{ \rm{ \:\displaystyle\int \rm \: {x}^{n}dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$