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Suppose the temperature in a region of space is given by $T(x,y,z) = 50z$. Find the average value of the temperature along the curve $C(t) = (t\cos(t) ,t\sin(t) , t)$ with $0 \le t\le 6$

I know that the average value is given by the equation $$ \overline{f}(C) = \frac{\int_Cf(x,y,z)ds}{\int_Cds}$$

I set up this equation like so:$$ \frac{\int_0^{6\pi}(50-t)\sqrt{(\cos(t)-t\sin(t))^2+(\sin(t)+t\cos(t))^2+1}dt}{\int_0^{6\pi}\sqrt{(\cos(t)-t\sin(t))^2+(\sin(t)+t\cos(t))^2+1}dt}$$

Which simplifies to $$ \frac{\int_0^{6\pi}(50-t)\sqrt{t^2+2}dt}{\int_0^{6\pi}\sqrt{t^2+2}dt}$$

Am I correct up to this point? These integrals are nasty ones, any help is appreciated.

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I didn't check your simplification, but assuming it is correct, it is now enough to compute $\int_0^{6 \pi} \sqrt{t^2+2} dt$ and $\int_0^{6 \pi} t \sqrt{t^2+2} dt$. The second is a routine $u$-substitution. The first one can be handled by trigonometric substitution, or (if you are familiar with it) hyperbolic substitution.

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