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I asked a question similar to this one, but I'm still confused on how to integrate this. I have $r'(t)=\langle 0,6t,3t^2\rangle$. and so this gives you the integral from $0$ to $\sqrt{12}$ of $\sqrt{36t^2+9t^4}dt$. Step by step would be helpful, thanks!

Wng427
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    To denote the three dimensional standard basis vectors I'd recommend using $\mathbf{\hat{i}},\mathbf{\hat{j}},$ and $\mathbf{\hat{k}}$. Codes are \mathbf{\hat{i}} etc. – K.defaoite Jun 03 '20 at 02:18

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You've set up the calculation to do, which is good. From here it is just calculus. If you factor out the $t^2$ from the square root, we end up having to compute

$$\int_{0}^{\sqrt{12}}t\sqrt{36 + 9t^2} \mathrm{d}t.$$

Do you see a substitution you could perform?

Oliver Daisey
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