Calculate the trajectory integral $$I=\int_{\gamma}zdl$$ where $\gamma$ is a curve obtained by the intersection of the surfaces given by the equations $x^2+y^2=z^2,y^2=ax$ from the point $(0,0,0)$ to $(a,a,\sqrt{2}a)$
** My attempt **
We need to parametrize the expression, note that the parametrization of $\gamma$ is given by $x=t$,$y=\sqrt{at}$,$z=\sqrt{t^2+at}$ where $t\in [0,a]$.
Now we find the derivatives of our parametrization $x^{\prime}=1$,$y^{\prime}=\frac{a}{2\sqrt{at}}$,$z^{\prime}=\frac{2t+a}{2 \sqrt{t^2+at}}$
now note that our parametrization is $C:[0,a]\rightarrow(t,\sqrt{at},\sqrt{t^2+at})$ where $C$ is injective trajectory, now we find $||C^{\prime}(t)||=\sqrt{(x^{\prime})^2+(y^{\prime})^2+(z^{\prime})^2}$ and $||C^{\prime}(t)||=\frac{1}{2}\sqrt{\frac{8t^2+9at+2a^2}{t^2+at}}$ now we substitute it in $I$ like a $$I=\int_{\gamma}zdl=\frac{1}{2}*\int_{0}^{a}\sqrt{t^2+at}\sqrt{\frac{8t^2+9at+2a^2}{t^2+at}}dt=\frac{1}{2}\int_{0}^{a} \sqrt{8t^2+9at+2a^2}dt$$ Finally solving the latest integral we get $$I=\int_{\gamma}zdl=\frac{1}{2}(\frac{1}{256}a\left(-72 \sqrt{2} \sqrt{a^2}+200 \sqrt{19} \sqrt{a^2}+17 \sqrt{2} a \log \left(8 \sqrt{a^2}+9 a\right)-17 \sqrt{2} a \log \left(4 \sqrt{38} \sqrt{a^2}+25 a\right)\right)$$ And like the intersection is two lines we multiply it by $2$ and we get $$I=\int_{\gamma}zdl=(\frac{1}{256}a\left(-72 \sqrt{2} \sqrt{a^2}+200 \sqrt{19} \sqrt{a^2}+17 \sqrt{2} a \log \left(8 \sqrt{a^2}+9 a\right)-17 \sqrt{2} a \log \left(4 \sqrt{38} \sqrt{a^2}+25 a\right)\right)$$ Is my approach fine? or probably I forget some detail.
I think that it wrong because usually when I apply a parametrization the integrand is much easy than it, and i had to compute the Integral in Wolphram Mathematica , because it is so much large. Please any help, hit or comment are useful and appreciate . Thanks so much for read me
Whichever way you look at it, it is not going to be a straightforward integral.
Please note the intersection curve is given by as below and we have to integrate over this curve from point $(0,0,0)$ to $(a,a,\sqrt2 \,a)$.
$x^2 + ax - z^2 = 0 \implies (x+\frac{a}{2})^2 - z^2 = \frac{a^2}{4}$
This is equation of a hyperbola and I would consider hyperbolic parametrization instead.
$x = -\frac{a}{2} + \frac{a}{2} \cosh t = \frac{a}{2} (\cosh t - 1) = a \sinh^2 \frac{t}{2}$
$z = \frac{a}{2} \, \sinh t$
$y = a \, \sinh \frac{t}{2}$
where $0 \leq t \leq \sinh^{-1}(2 \sqrt2)$ (bound based on $0 \leq z \leq \sqrt 2 \, a$).
So $r(t) = (a \sinh^2 \frac{t}{2}, a \sinh \frac{t}{2}, \frac{a}{2} \, \sinh t)$
$r'(t) = (a \sinh \frac{t}{2} .\cosh \frac{t}{2}, \frac{a}{2} \, \cosh \frac{t}{2}, \frac{a}{2} \, \cosh t)$
$||r'(t)|| = \displaystyle \frac{a \sqrt {2 \cosh 2t + \cosh t + 1}}{2 \sqrt2}$
$I = \displaystyle \frac{a^2}{4 \sqrt2} \int_0^{\sinh^{-1}(2 \sqrt2)} (\sinh t \sqrt {2 \cosh 2t + \cosh t + 1}) \, dt = \frac{a^2}{4 \sqrt2} \times 8.22216 \approx 1.4535a^2$