[Solved] The Area Between The Parabolas Y2 = 4ax And X2 = 4ay Is

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y2 = 4ax & x2 = 4ay

F1 U.B Madhu 28.12.19 D 3

We have to find shaded region area.

So area drawn by \(y = \frac{{{x^2}}}{{4a}}\) on x-axis = A1 (say)

Then shaded area = |A1 – A2|

So, \({A_1} = \mathop \smallint \nolimits_0^{4a} y \cdot dx = \mathop \smallint \nolimits_0^{4a} \frac{{{x^2}}}{{4a}} \cdot dx = \left[ {\frac{{{x^3}}}{{12a}}} \right]_0^{4a} = \frac{{16{a^2}}}{3}\)

\({A_2} = \mathop \smallint \nolimits_0^{4a} y \cdot dx = \mathop \smallint \nolimits_0^{4a} \sqrt {4ax \cdot dx} = \sqrt {4a} \cdot \left[ {\frac{{{{\left( x \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^{4a} = \frac{{4\sqrt a }}{3}\;\left[ {{{\left( x \right)}^{\frac{3}{2}}}} \right]_0^{4a}\)

\(\Rightarrow {A_2} = \frac{{32{a^2}}}{3}\)

⇒ Shaded area \(= \left| {{A_1} - {A_2}} \right| = \left| {\frac{{16{a^2}}}{3} - \frac{{32{a^2}}}{3}} \right| = \frac{{16{a^2}}}{3}\) Download Solution PDF Share on Whatsapp

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