Question

Find the area bounded between

${x}^{2} + {y}^{2} = {d}^{2} \: and \: x + y = d$

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Answer 1

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

Given curves are

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

and

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

Step :- 1 Point of intersection

From equation (2), we have

$\rm :\longmapsto\:y = d - x$

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

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

$\rm :\longmapsto\: {x}^{2} + {d}^{2} + {x}^{2} - 2dx= {d}^{2}$

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

$\rm :\longmapsto\:2x(x - d) = 0$

$\bf\implies \:x = 0 \: \: \: or \: \: \: x = d$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

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

Step :- 2 Curve Sketching

The curve

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

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

[ See the attachment ]

Step :- 3 Required area bounded between Curves

Required area bounded between curves is

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

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

$\rm \:  =  \:\displaystyle\int_{0}^{d}\bigg[ \sqrt{ {d}^{2} - {x}^{2}} - d + x \bigg] \: dx$

As we know that,

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

So, using this, we get

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

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

$\rm \:  =  \: \dfrac{ {d}^{2} }{2} {sin}^{ - 1}1 - \dfrac{ {d}^{2} }{2}$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2} \times \dfrac{\pi}{2} - \dfrac{ {d}^{2} }{2}$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2}\bigg[ \dfrac{\pi}{2} -1\bigg]$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2}\bigg[ \dfrac{\pi - 2}{2}\bigg]$

$\rm \:  =  \: \dfrac{ {d}^{2}(\pi - 2) }{4} \: square \: units$

Answer 2

Answer:

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

Given curves are

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

and

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

Step :- 1 Point of intersection

From equation (2), we have

$\rm :\longmapsto\:y = d - x$

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

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

$\rm :\longmapsto\: {x}^{2} + {d}^{2} + {x}^{2} - 2dx= {d}^{2}$

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

$\rm :\longmapsto\:2x(x - d) = 0$

$\bf\implies \:x = 0 \: \: \: or \: \: \: x = d$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

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

Step :- 2 Curve Sketching

The curve

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

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

[ See the attachment ]

Step :- 3 Required area bounded between Curves

Required area bounded between curves is

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

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

$\rm \:  =  \:\displaystyle\int_{0}^{d}\bigg[ \sqrt{ {d}^{2} - {x}^{2}} - d + x \bigg] \: dx$

As we know that,

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

So, using this, we get

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

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

$\rm \:  =  \: \dfrac{ {d}^{2} }{2} {sin}^{ - 1}1 - \dfrac{ {d}^{2} }{2}$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2} \times \dfrac{\pi}{2} - \dfrac{ {d}^{2} }{2}$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2}\bigg[ \dfrac{\pi}{2} -1\bigg]$

$\rm \:  =  \: \dfrac{ {d}^{2} }{2}\bigg[ \dfrac{\pi - 2}{2}\bigg]$

$\rm \:  =  \: \dfrac{ {d}^{2}(\pi - 2) }{4} \: square \: units$