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In case you guys forgot, the formula for arc length is: $$\int_a^b\sqrt{1+(\frac{dx}{dy})^2}dy$$ So my integral is: $$\int_{-27}^{343}\sqrt{1 + (4y^\frac13-\frac{1}{16}y^\frac{-1}{3})^2}dy$$ $$= \int_{-27}^{343}\sqrt{1 + 16y^\frac19-\frac12+\frac{1}{256}y^\frac{-1}{9}}dy$$ $$= \int_{-27}^{343}4y^\frac13+\frac{1}{16}y^\frac{-1}{3}dy$$ $$= 3y^\frac43+\frac{3}{32}y^\frac23\;from\;y = -27\;to\;y = 343$$ $$= 3(343^\frac43-(-27)^\frac43) + \frac{3}{32}(343^\frac23 - (-27)^\frac23)$$ $$= 6960 + \frac{15}{4} = 6963.75$$

But somehow this answer is wrong. I'm doing an online homework for my Calculus 2 class and I really don't know what did I do wrong. Please help!

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2 Answers 2

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$(y^{\frac 13})^2=y^{\frac 23}$, not $y^{\frac 19}$, but that error is neutralized the next line. You lost a minus sign integrating $y^{\frac{-1}3}$, but that reduces the answer. You should be integrating the absolute value, so need to break it at zero. You have a negative integral when $y \lt 0$. Alpha gets $7451.4375$

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  • $\begingroup$ To make the algegra a bit simpler, express the curve as $y=t^3$, $x=3t^4-\frac{3}{32}t^2$ for $-3\le t\le 7$. The arclength (using the formula for parametric equations) is $$\int_{-3}^7\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int_{-3}^7\sqrt{\left(3t^2\right)^2+\left(12t^3-\frac{3t^2}{16}\right)^2}\,dt=\int_{-3}^7\left\vert12t^3+\frac{3}{16}t\,\right\vert\,dt$$ $\endgroup$
    – Steve Kass
    Commented Nov 30, 2014 at 6:43
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Note $\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}$ is not defined for $y=0$, then we must take $$\text{Arc length}=\int_{-27}^0\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy+\int_{0}^{343}\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy$$

Where $$\int_{-27}^0\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy=\lim_{b\to0^-}\int_{-27}^b\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy$$ and $$\int_{0}^{343}\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy=\lim_{a\to0^+}\int_{a}^{343}\sqrt{1+\left(\frac{\mathrm d x}{\mathrm d y}\right)^2}\mathrm dy$$

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