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**Showcase_22** Evaluate the line integral $\displaystyle \int_C(x+y^2)$, where $\displaystyle C$ is the parabola $\displaystyle y=x^2$ in the plane $\displaystyle z=0$ connecting the points $\displaystyle (0,0,0)$ and $\displaystyle (2,4,0)$.

My method:

Okay, so it's probably good to start with a parameterisation:

$\displaystyle \gamma (t)=(2t, 4t^2,0)$ where $\displaystyle t \in [0,1]$.

I need to find $\displaystyle \int_0^1 f(\gamma (t)) || \gamma (t) ||~dt$ where $\displaystyle f(x,y)=x+y^2$

== Probably a typo: if must be $\displaystyle || \gamma^{'} (t)||$ in the integral

$\displaystyle \int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt$

$\displaystyle \int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt$

$\displaystyle 2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt$

$\displaystyle 4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the first integral:

Let $\displaystyle u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t$

$\displaystyle 4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du$

$\displaystyle \frac{1}{8} \left[ \frac{3}{2} u^{\frac{2}{3}} \right]_1^{17}$

$\displaystyle =\frac{3}{16}(17^{\frac{2}{3}}-1)$

=== This is wrong: $\displaystyle \int{\sqrt{u} \, du} = \frac{2}{3}u^{\frac{3}{2}}$ and not what you wrote

Tonio

So the two integrals are now:

$\displaystyle \frac{3}{16}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the second integral:

Let $\displaystyle (4t)^2=\cosh 2u$

Then, for the second integral, we have $\displaystyle 32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du$

Which becomes:

$\displaystyle \frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du$

by FTC, the integral is $\displaystyle \left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}$

so substituting the limits gives:

$\displaystyle \frac{1}{18} (16)^{\frac{9}{2}}$

So my overall answer to the question is:

$\displaystyle \frac{3}{16}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}$

Firstly, is this right? I have an $\displaystyle arcosh 0$ in there that is undefined, but it removes itself quite nicely at the end.

Secondly, is there a faster way of doing this???