Math, asked by karansinghsaggu, 4 days ago

Find the area of circle x^२ + y^२ =४ using integration​

Answers

Answered by mayanknagdali122993
0

Answer:

Equation of the circle is x

2

+y

2

=4

∴y=

4−x

2

But the area element is given by

∴ydx=

4−x

2

dx

Integrating both the sides with x ranging from -2 to 2

∴A=∫

−2

2

ydx=∫

−2

2

4−x

2

dx

Let x=2sinθ ⇒dx=2cosθdθ

∴A=∫

−π

π

4−4sin

2

θ

2cosθdθ=∫

−π

π

4cos

2

θdθ

=2∫

−π

π

(1+cos2θ)dθ

=4π

Answered by mathdude500
8

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

Given equation of circle is

\rm \:  {x}^{2} +  {y}^{2} = 4 \\

can be rewritten as

\rm \:  {(x - 0)}^{2} +  {(y - 0)}^{2} =  {2}^{2}  \\

It means, circle having centre (0, 0) and radius, r = 2 units.

Now, see the attachment.

We concluded that circle divided in to 4 equal parts in 4 quadrants, So we find the area in first quadrant with respect to x - axis from x = 0 to x = 2 and multiply by 4 to find the required area.

So, required area of circle is

\rm \:  =  \:4 \:  \displaystyle\int_{0}^{2}\rm y \: dx \\

\rm \:  =  \:4 \:  \displaystyle\int_{0}^{2}\rm  \sqrt{4 -  {x}^{2} }  \: dx \\

\rm \:  =  \:4 \:  \displaystyle\int_{0}^{2}\rm  \sqrt{ {2}^{2}  -  {x}^{2} }  \: dx \\

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

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

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

\rm \:  =  \:4 \times   2{sin}^{ - 1} 1 \\

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

\rm \:  =  \: 4\pi \: square \: units \\

\rule{190pt}{2pt}

Formulae Used :-

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

\rule{190pt}{2pt}

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}

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