Arc length of $\alpha(t)=(t^6,t^9)$

Question

I have to evaluate the arc length of the curve $$\alpha(t)=(t^6,t^9)$$ for $t\in[-\sqrt[3]{2},\sqrt[3]{2}]$. But when I do it, the final integral is $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$ which value is zero. What should I do?

Answer 1

Remember that $\sqrt{x^2}=|x|$.

So as mentioned by Arctic Char it should be $|t|^5$. Then you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt=$$ $$=\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt-\int_{-\sqrt[3]{2}}^{0}3t^{5}\sqrt{4+9t^{6}}dt=$$ $$=\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=0}^{t=\sqrt[3]{2}}-\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=-\sqrt[3]{2}}^{t=0}$$ $$=\frac{1}{27}[40^{\frac{3}{2}}-8]-\frac{1}{27}[8-40^{\frac{3}{2}}]$$ $$=\frac{1}{27}[160\sqrt{10}-16]=\frac{16}{27}[10\sqrt{10}-1] \space(\approx18.14683)$$

(where $40^{\frac{3}{2}}=40\sqrt{40}=80\sqrt{10}$).

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Or since the integrand is an even function you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt$$ $$=2\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$ $$=2\cdot\frac{1}{27}[80\sqrt{10}-8]=\frac{16}{27}[10\sqrt{10}-1].$$