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$$r = 2\theta^2, 0<=\theta <=\sqrt{5}$$ calc the length of the curve.

Since it's probably polar coordinats the formulat should be:

$$\int_0^\sqrt{5} \sqrt{(r(\theta))^2+(r'(\theta))^2} d\theta = \int_0^\sqrt{5} \sqrt{(2\theta^2)^2+(4\theta)^2} d\theta $$. Wolfram alpha gives the anser 18 for this but the correct answer should be $\frac{38}{3}$ so I suppose my approach is wrong. What's the error?

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This is the correct approach. You must have mistyped it in WolframAlpha because it returns the correct answer $\frac{38}{3}$. Make sure that you input the parentheses as they should be. In general, the arc length of a polar curve $r=r(\theta)$ over an interval $(\alpha,\beta)$ is given by $$\mathscr{L}=\int_{\alpha}^{\beta} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta\,,$$ which is exactly the expression that you used in your calculation (sometimes the best approach in computing this integral is to revert to rectangular coordinates, however).

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