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Prove that the curves $y^2 = 4x$ and $x^2 = 4y$ Prove that the curves $x = 0, x= 4, y= 0 , y =4$ into three equal parts.

Answers (1)

Given: $y^2 = 4x\qquad -(i)$

$x^2 = 4y\qquad -(ii)$

$x = 0, x= 4, y= 0 , y =4$

For point of intersection put $y = \frac{x^2}{4}$ in (i)

$\\\Rightarrow \left(\frac{x^2}{4} \right )^2 = 4x \Rightarrow \frac{x^4}{16} = 4x\\ \Rightarrow x^3 = 4 \text{ or } x^4 =64 x \\\Rightarrow x(x^3-64) =0\qquad x = 0, x= 4$

Area of Part A,

$I = \int_0^4x \text{ of the curve I}dy$

$I = \int_0^4\frac{y^4}{4}dy$

$\Rightarrow \frac{1}{4\times 3}[y^3] \Rightarrow \frac{1}{12} [ 4^3]$

$I = \frac{64}{12} \Rightarrow \frac{16}{3} \text{ sq.units}$

Area of Part B,

$I = \int_0^4 y\text{ of the 1}^{st}\text{ curve}dx - \int_0^4 y\text{ of the 2}^{nd}\text{ curve}dx$

$\Rightarrow \int_0^42\sqrt x dx - \int_0^4\frac{x^2}{4}dx$

$\Rightarrow \frac{2\times 2}{3}\left[x^{\frac{3}{2}} \right ]_0^4 - \frac{1}{12}\left[x^3 \right ]_0^4$

$I \Rightarrow \frac{4}{3}\left[4^{\frac{3}{2}} \right ] - \frac{1}{12}\left[4^3 \right ]$

$\Rightarrow \frac{4}{3}\times 8 - \frac{1}{12}\times 64\Rightarrow \frac{32}{3} - \frac{16}{3}$

$I = \frac{16}{3} \text{ sq.units}$

Area of part C

$I = \int_0^4y \text{ of the curve I}dx$

$I = \int_0^4\frac{x^4}{4}dx$

$\Rightarrow \frac{1}{4\times 3}[x^3] \Rightarrow \frac{1}{12} [ 4^3]$

$I = \frac{64}{12} \Rightarrow \frac{16}{3} \text{ sq.units}$

$I = \frac{16}{3} \text{ sq.units}$

Posted by

Ravindra Pindel

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