Question

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Please don't respond if you can't explain how to get out of an impossible square root. Thanks

Answer 1

Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
$\int_a^b f(r) |r'| dt$

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector $\sqrt{(x')^2 + (y')^2 + (z')^2}$

$\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt$

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3  dt

$\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du$

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
$=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\ \\ =\frac{1}{6} (14^{3/2} - 1)$