I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = \sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 \leq x \leq 13$
I start by parameterizing the curve and the surface for $0 \leq t \leq 13:
$\begin{align} r(t) &= \langle t, t^{3/2} \rangle \\ f(t) &= \sqrt{16+36t} \\ \end{align}$
And calculate a few things I'll need later:
$\begin{align} r'(t) &= \langle 1, \frac{3}{2}t^{1/2} \rangle \\ \left|r'(t)\right| &= \sqrt{ 1 + \frac{9}{4}t } \\ \end{align}$
Next, I calculate the line integral:
$\begin{align} &\int_0^{13} f(t) \left|r'(t)\right| dt \\ &\int_0^{13} \sqrt{16+36t} \sqrt{ 1 + \frac{9}{4}t }\ dt \\ 2 &\int_0^{13} \sqrt{4+9t} \sqrt{ 1 + \frac{9}{4}t }\ dt \\ 2 &\int_0^{13} \sqrt{ \frac{4}{4} (4+9t) } \sqrt{ 1 + \frac{9}{4}t }\ dt \\ 4 &\int_0^{13} \sqrt{ 1 + \frac{9}{4}t } \sqrt{ 1 + \frac{9}{4}t }\ dt \\ 4 &\int_0^{13} 1 + \frac{9}{4}t\ dt \\ 4 &\left( t + \frac{9}{8}t^2\right|_0^{13} \\ 4 &\left( 13 + \frac{9}{8}13^2\right) \\ &\frac{1625}{2} \\ \end{align}$
Then, I calculate the length of the curve:
$\begin{align} L &= \int_0^{13} \sqrt{1 + [r'(t)]^2}\ dx \\ L &= \int_0^{13} \sqrt{1 + 1 + \frac{9}{4}t}\ dx \\ L &= \int_0^{13} \sqrt{2 + \frac{9}{4}t}\ dx \\ \end{align}$
Letting $u = 2 + 9/4 t$
$\begin{align} L &= \frac{4}{9} \int_2^{125\ /\ 4} u^{1/2}\ du \\ L &= \frac{8}{27} \left( u^{3/2} \right|_{\ 2}^{\ 125\ /\ 4} \\ L &= \frac{8}{27} \left[ \left(\frac{125}{4}\right)^{3/2} - 2^{3/2} \right] \\ L &= \frac{625 \sqrt{5} - 16 \sqrt{2} }{27} \\ \end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
you mess with the integral of the curve, have mix dx with t. $$r'(t)=\frac {dr}{dt}=\hat T \frac {ds}{dt}$$ $$ds=\sqrt{1+\frac 94 t^2}dt$$ That is easy solve with trig substitution. $$\int_0^{13} \sqrt{1+\frac 94 t^2}dt=128.1379$$ $$ ave(f(x,y))=\frac {1625}{2\cdot 128.1379}= 6.3408 $$