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this is my curve : $r(t)=(\cos{t},\sin{t}-1,2\cos{\frac{t}{2}})$ , $t=[0,3\pi]$

$r'(t)=(-\sin{t},\cos{t},-\sin{t})$

$||r'(t)||=\sqrt{\sin^2{t}+1}$

I have to calculate: $\int{(y+1)}ds$

So I have : $\int_{0}^{3\pi}\sin t\sqrt{\sin^2{t}+1}dt$

Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.

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    $\begingroup$ You have a mistake in differentiation of $\cos\frac{t}{2}$ (you skipped the 1/2 in $\sin$). $\endgroup$ – orion Jan 19 at 23:43
  • $\begingroup$ @orion it’s $2\cos\frac{t}{2}$, so he doesn’t. $\endgroup$ – cluelessatthis Jan 20 at 0:45
  • $\begingroup$ It should be $-\sin \frac{t}{2}$, not $-\sin t$. The half doesn't disappear from inside the trigonometric function after differentiation. $\endgroup$ – orion Jan 20 at 0:52
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$r(t)=(\cos{t},\sin{t}-1,2\cos{\frac{t}{2}})$ , $t=[0,3\pi]$

$r'(t)=(-\sin{t},\cos{t},-\sin{\frac{t}{2}})$ <- You forgot sin(t/2)

$||r'(t)||=\sqrt{\sin^2{\frac{t}{2}}+1}$

Calculate: $\int{(y+1)}ds$

-> $I = \int_{0}^{3\pi}\sin t\sqrt{\sin^2{\frac{t}{2}}+1}dt$

Apply subsitution $u = sin^2(\frac{t}{2}) + 1$

$du = \frac{\sin{t}}{2} dt$

$t = 0, u = 1. t = 3\pi,u = 2.$

$I = 2\int_{1}^{2}\sqrt{u}\textrm{ }du = \frac{4}{3}(2^\frac{3}{2}-1^\frac{3}{2})=\frac{4}{3}\times(2\sqrt{2}-1)=\frac{8\sqrt{2}-4}{3}$

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  • $\begingroup$ why is It $du=\frac{sint}{2}$ ? how did you get $\frac{sint}{2}$? $\endgroup$ – NPLS Jan 20 at 8:44
  • $\begingroup$ $u = sin^2(\frac{t}{2}); du = 2sin(\frac{t}{2})cos(\frac{t}{2})\times \frac{1}{2}$ By Chain Rule. Simplify. $\endgroup$ – Gareth Ma Jan 21 at 9:05

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