Question

Find the smallest area bounded between the curve x^2 + y^2 = 1 and x + y = 1

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Answer 1

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

Given curves are

$\rm \: {x}^{2} + {y}^{2} = 1$

and

$\rm \: x + y = 1$

Step :- 1 Point of intersection of two curves

As, it is given that,

$\rm \: x + y = 1$

$\rm\implies \:y = 1 - x$

On substituting the value of y in second curve, we get

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

$\rm \: {x}^{2} + {x}^{2} + 1 - 2x = 1$

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

$\rm \: 2x(x - 1) = 0$

$\rm\implies \:x = 0 \: \: or \: \: x = 1$

Hence, the point of intersection of two curves are as

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

Step :- 2 Curve Sketching

$\rm \: {x}^{2} + {y}^{2} = 1$

represents the equation of circle having center (0, 0) and radius 1 unit.

and

x + y = 1 represents the equation of line which passes through the point (0, 1) and (1, 0).

[ See the graph in attachment ]

Step :- 3 Required Area

As we have to find the smallest area bounded between these two curves, so shaded region represents the required area.

So, required area between the two curves is

$\rm \: = \: \displaystyle\int_0^1\rm (y_{circle} \: - \: y_{line} \: ) \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm ( \sqrt{1 - {x}^{2} } - (1 - x) \: ) \: dx$

$\rm \: = \: \bigg[\dfrac{x}{2} \sqrt{1 - {x}^{2}} + \dfrac{1}{2} {sin}^{ - 1}x - x + \dfrac{ {x}^{2} }{2} \bigg]_0^1$

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

$\rm \: = \: \dfrac{1}{2} \times \dfrac{\pi}{2} - 1 + \dfrac{ 1 }{2}$

$\rm \: = \: \dfrac{\pi}{4} - \dfrac{ 1 }{2}$

$\rm \: = \: \dfrac{\pi - 2}{4} \: square \: units$

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Formula Used

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

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

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Additional Information

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$