Question

Find the area bounded between the curves y²= 4ax and x²= 4by

Answer 1

Answer

16ab is the answer

_____ square units

3

Answer 2

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

The given curves are

$\rm :\longmapsto\: {y}^{2} = 4ax - - - (1)$

and

$\rm :\longmapsto\: {x}^{2} = 4by - - - - (2) \:$

Let first find the point of intersection of two curves.

From equation (2), we have

$\rm :\longmapsto\: {x}^{2} = 4by$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{4} = 16 {b}^{2} {y}^{2}$

can be rewritten as using equation (1),

$\rm :\longmapsto\: {x}^{4} = 16 {b}^{2} (4ax)$

$\rm :\longmapsto\: {x}^{4} - 64 {ab}^{2}x = 0$

$\rm :\longmapsto\:x( {x}^{3} - 64 {ab}^{2}) = 0$

$\rm \implies\:x = 0 \: \: or \: {x}^{3} = 64 {ab}^{2}$

$\rm \implies\:x = 0 \: \: or \: x = 4( {ab}^{2})^{ \frac{1}{3} }$

See the attachment for the required area to be evaluated.

Now, from equation (1)

$\rm :\longmapsto\: {y}^{2} = 4ax$

$\red{\rm :\longmapsto\:y_1 = 2 \sqrt{a} \sqrt{x} }$

And from equation (2), we have

$\rm :\longmapsto\: {x}^{2} = 4by$

$\red{\rm :\longmapsto\:y_2 = \dfrac{ {x}^{2} }{4b}}$

So, required area between two curves is given by

$\rm \:  =  \: \displaystyle\int_0^{ 4( {ab}^{2})^{ \frac{1}{3}}}\bigg[y_1 - y_2\bigg] \: dx$

$\rm \:  =  \: \displaystyle\int_0^{ 4( {ab}^{2})^{ \frac{1}{3}}}\bigg[2 \sqrt{a} \sqrt{x} - \frac{ {x}^{2} }{4b} \bigg] \: dx$

$\rm \:  =  \: 2 \sqrt{a} \times \dfrac{2}{3} \bigg[ {x}^{ \frac{3}{2} } \bigg] _0^{ 4( {ab}^{2})^{ \frac{1}{3} } } - \dfrac{1}{4b}\bigg[\dfrac{ {x}^{3} }{3} \bigg]_0^{ 4( {ab}^{2})^{ \frac{1}{3} } }$

$\rm \:  =  \: \dfrac{32\sqrt{a} }{3} \sqrt{a {b}^{2} } \: - \dfrac{1}{12b} \times 64 {( {ab}^{2} )}$

$\rm \:  =  \: \dfrac{32ab}{3} - \dfrac{16ab}{3}$

$\rm \:  =  \: \dfrac{16ab}{3} \: sq. \: units$

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Additional Information

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