CurvesHard

Line integral — ∫_γ (x+y) ds

Calculate the line integral

\int_\gamma(x+y)\,ds

on γ(t)=(t,t),

t\in[0,1]

Given data

γ(t) = (t, t)t∈[0,1]

Steps

0 / 3

Calculate |γ'(t)|.
Write your working — the numerical result must appear in the text (logica)
Substitute f(γ(t)) = t+t.
Calculate ∫₀¹ 2t·√2 dt.

Full worked solution

Calculate |γ'(t)|.
$|\gamma'| = \sqrt{1^2+1^2} = \sqrt2$
$\gamma'(t) = (1, 1)$ . The magnitude is $|\gamma'(t)| = \sqrt{1^2 + 1^2} = \mathbf{\sqrt{2}}$ . The arc length element is $ds = |\gamma'(t)|\,dt = \sqrt{2}\,dt$ .
Substitute f(γ(t)) = t+t.
$f = 2t$
Along $\gamma$ , $x = t$ , $y = t$ , so $f(\gamma(t)) = x + y = t + t = \mathbf{2t}$ . The integrand expressed in terms of the parameter $t$ .
Calculate ∫₀¹ 2t·√2 dt.
$\sqrt2\cdot[t^2]_0^1 = \sqrt2 \approx 1.414$
$\int_\gamma (x+y)\,ds = \int_0^1 2t \cdot \sqrt{2}\,dt = \sqrt{2} \int_0^1 2t\,dt = \sqrt{2} \cdot [t^2]_0^1 = \sqrt{2} \approx \mathbf{1.414}$ .

Result:∫_γ(x+y) ds = √2 ≈ 1.414.

Keep studying