Question

find the length of the arc of the curve 9y²= (x²+2)³ from the point where x=0 To the point where x = 2.

Answer 1

$$3y=\pm(x^2+2)^{3/2}$$

$$y=\pm\dfrac{1}{3}(x^2+2)^{3/2}$$

Take $y=\dfrac{1}{3}(x^2+2)^{3/2}$

$$y'=\dfrac{1}{3}(\dfrac{3}{2})(x^2+2)^{1/2}(2x)=x\sqrt{x^2+2}$$

$$L=2\displaystyle\int_{0}^{2}\sqrt{1+(x\sqrt{x^2+2})^2}dx$$

$$=2\displaystyle\int_{0}^{2}\sqrt{1+2x^2+x^4}dx$$

$$=2\displaystyle\int_{0}^{2}\sqrt{(1+x^2)^2}dx$$

$$=2\displaystyle\int_{0}^{2}(1+x^2)dx$$

$$=2[x+\dfrac{x^3}{3}]\begin{matrix} 2 \\ 0 \end{matrix}=2(2+\dfrac{8}{3}-0)=\dfrac{28}{3}$$