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Question

# Evaluate $\int_{\mathbf{c}} f\, ds$, where $f (x, y, z) = z$ and $\mathbf{c}(t) = (t \cos t, t \sin t, t)$ for $0 \leq t \leq t_0$.

Solution

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Curve $\mathbf{c}$ is parametrized by

\begin{equation*} \left\{\begin{aligned} x(t)&=t\cos t,\\ y(t)&=t\sin t,\\ z(t)&=t, \end{aligned}\right.\quad 0\leq t\leq t_0. \end{equation*}

First, we compute $||\mathbf{c}'(t)||$:

\begin{align*} ||\mathbf{c}'(t)||&= \sqrt{\left[\dfrac{dx(t)}{dt}\right]^2+\left[\dfrac{dy(t)}{dt}\right]^2+\left[\dfrac{dz(t)}{dt}\right]^2}\\&=\sqrt{\left [\dfrac{d(t\cos t)}{dt}\right ]^2+\left [\dfrac{d(t\sin t)}{dt}\right ]^2+\left [\dfrac{dt}{dt}\right ]^2}\\ &=\sqrt{(\cos t-t\sin t)^2+(\sin t+t\cos t)^2+1^2}\\ &=\sqrt{\cos^2 t+\sin^2 t+t^2(\cos^2 t+\sin^2t)+1}\\ &=\sqrt{1+t^2+1}\\ &=\sqrt{t^2+2}. \end{align*}

Next, we substitute for $x$, $y$, $z$ in terms of $t$ to obtain

$\begin{equation*} f(x,y,z)=z=t, \end{equation*}$

along $\mathbf{c}$.

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