Answer to Question #347441 in Calculus for harry

Question #347441

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.


1
Expert's answer
2022-06-07T08:46:14-0400
3y=±(x2+2)3/23y=\pm(x^2+2)^{3/2}

y=±13(x2+2)3/2y=\pm\dfrac{1}{3}(x^2+2)^{3/2}

Take y=13(x2+2)3/2y=\dfrac{1}{3}(x^2+2)^{3/2}


y=13(32)(x2+2)1/2(2x)=xx2+2y'=\dfrac{1}{3}(\dfrac{3}{2})(x^2+2)^{1/2}(2x)=x\sqrt{x^2+2}


L=2021+(xx2+2)2dxL=2\displaystyle\int_{0}^{2}\sqrt{1+(x\sqrt{x^2+2})^2}dx

=2021+2x2+x4dx=2\displaystyle\int_{0}^{2}\sqrt{1+2x^2+x^4}dx

=202(1+x2)2dx=2\displaystyle\int_{0}^{2}\sqrt{(1+x^2)^2}dx

=202(1+x2)dx=2\displaystyle\int_{0}^{2}(1+x^2)dx

=2[x+x33]20=2(2+830)=283=2[x+\dfrac{x^3}{3}]\begin{matrix} 2 \\ 0 \end{matrix}=2(2+\dfrac{8}{3}-0)=\dfrac{28}{3}


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