<s>[INST] Average Value of a Line Integral 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?
 [/INST] 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.
$$\int_0^{13} \sqrt{1+\frac 94 t^2}dt=128.1379$$
$$ ave(f(x,y))=\frac {1625}{2\cdot 128.1379}= 6.3408 $$
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