Question

Find the area between the curves $\rm {x}^{2} + {y}^{2} = 16$ and $\rm {(x + 4)}^{2} + {y}^{2} = 16.$

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Answer 1

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

Given curves are

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

and

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

Step :- 1 Point of intersection

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

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

$\rm \: 8x + 16 = 0$

$\rm \: 8x = - 16$

$\rm\implies \:x = - 2$

So,

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

$\rm \: 4+ {y}^{2} = 16$

$\rm \: {y}^{2} = 16 - 4$

$\rm \: {y}^{2} = 12$

$\rm\implies \:y \: = \: \pm \: 2 \sqrt{3}$

Hᴇɴᴄᴇ,

➢ Pair of points of intersecting of two given curves are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 2 & \sf 2 \sqrt{3} \\ \\ \sf - 2 & \sf - 2 \sqrt{3} \end{array}} \\ \end{gathered}$

Step :- 2 Curve Sketching

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

represents a circle having center (0, 0) and radius 4 units

Let

$\rm \: y_1 = \sqrt{16 - {x}^{2} }$

And

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

represents a circle having center (- 4, 0) and radius 4 units.

$\rm \: y_2 = \sqrt{16 - {(x + 4)}^{2} }$

Step :- 3 Required Area

So, required area between these two curves with respect to x - axis is given by

$\rm \: =2\bigg( \displaystyle\int_{ - 4}^{ - 2}\rm y_1 \: dx \: + \displaystyle\int_{ - 2}^{0}\rm y_2 \: dx\bigg)$

Now, Consider first

$\displaystyle\int_{ - 2}^{0}\rm y_2 \: dx$

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

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

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

$\rm \: = 4\pi - \bigg[\dfrac{ - 2 + 4}{2} \sqrt{16 - {( - 2 + 4)}^{2} } + 8{sin}^{ - 1} \dfrac{ - 2 + 4}{4} - 0 - 0\bigg]$

$\rm \: =4\pi- \sqrt{16 - 4}+ 8{sin}^{ - 1} \dfrac{1}{2}$

$\rm \: = 4\pi - \sqrt{12}-8 \times \dfrac{\pi}{6}$

$\rm \: = -2 \sqrt{3}+\dfrac{8\pi}{3} \: sq. \: units$

Now, Consider

$\rm \: \displaystyle\int_{ - 4}^{-2}\rm y_1 \: dx$

$\rm \: = \displaystyle\int_{ - 4}^{-2}\rm \sqrt{16 - {x}^{2} } \: dx$

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

$\rm \: =\rm \: \bigg[ \dfrac{x}{2} \sqrt{16 - {x}^{2} } + \dfrac{16}{2} {sin}^{ - 1} \dfrac{x}{4}\bigg]_{ - 4}^{-2}$

$\rm \: =\rm \: \bigg[ \dfrac{ - 2}{2} \sqrt{16 - 4 } + 8 {sin}^{ - 1} \dfrac{ - 2}{4}\bigg] - (-4\pi )$

$\rm \: =\rm \: 4\pi - \sqrt{12 } + 8 {sin}^{ - 1} \dfrac{1}{2}$

$\rm \: = 4\pi -2 \sqrt{3} - 8 \times \frac{\pi}{6}$

$\rm \: = - 2 \sqrt{3} + \frac{8\pi}{3} \: sq. \: units$

So, on substituting these values in

$\rm \: =2\bigg( \displaystyle\int_{ - 4}^{ - 2}\rm y_1 \: dx \: + \displaystyle\int_{ - 2}^{0}\rm y_2 \: dx\bigg)$

$\rm \: =2 \bigg(-2 \sqrt{3} + \frac{8\pi}{3} - 2 \sqrt{3} + \frac{8\pi}{3}\bigg) \: sq. \: units$

$\rm \: =2 \bigg(-4\sqrt{3} + \frac{16\pi}{3}\bigg) \: sq. \: units$

$\rm \: =-8\sqrt{3} + \frac{32\pi}{3} \: sq. \: units$

Hence,

$\rm\implies \:\rm \: Required\:Area =-8\sqrt{3} + \frac{32\pi}{3} \: sq. \: units$

Answer 2

Answer:

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