Consider the curve
\gamma: [0,L] \to \mathbb R^3,
\gamma(t)
= { \color{teal}\begin{pmatrix} x(t) \\y(t) \\z (t) \end{pmatrix}
= \begin{pmatrix}
A \cdot t^2 + B\cdot t \\
C \cdot t + D \\
E \cdot t
\end{pmatrix}}
and the vector field
K: \mathbb R^3 \to \mathbb R^3 defiined by
\color{purple}K(x,y,z) =\begin{pmatrix}
X z \\
y \\
Z x \end{pmatrix}.
Compute
\displaystyle \int_\gamma K d\gamma.
\displaystyle \int_\gamma K d\gamma=
By definition the integral \displaystyle \int_\gamma K d\gamma for
\gamma: [0,{\mathbfL}] \to \mathbb R^3 is
\displaystyle \int_0^{{\mathbfL}} {\color{red}K(\gamma(t))} \cdot {\color{blue}\gamma'(t)} dt.
The (velocity-)vector is { \color{blue}\gamma'(t)
= \begin{pmatrix} \left(A \cdot t^2 + B \cdot t\right)' \\
\left(C \cdot t + D \right)' \\
\left(E \cdot t\right)'
\end{pmatrix}
= \begin{pmatrix} 2*A \cdot t + B \\
C \\ E
\end{pmatrix}}.
The curve substitute into the vector field gives the 2nd factor for the scalar product
\color{red}K(\gamma(t)) =\begin{pmatrix} X z(t) \\ y(t) \\
Z x(t) \end{pmatrix} =
\begin{pmatrix} X \left (E t \right) \\ C \cdot t + D \\
Z \left(A \cdot t^2 + B\cdot t \right) \end{pmatrix}
=
\begin{pmatrix} X*E t \\ C \cdot t + D \\
Z*A \cdot t^2 + Z*B\cdot t \end{pmatrix}.
Therefore we get
{\color{red}\begin{pmatrix} X*E t \\ C \cdot t + D \\
Z*A \cdot t^2 + Z*B\cdot t \end{pmatrix}} \cdot
{\color{blue}\begin{pmatrix} 2*A \cdot t + B \\
C \\ E\end{pmatrix}} =
\left( X*E t \right ) \cdot \left( 2*A t + B \right ) +
(C \cdot t + D) \cdot negParens(C) +
\left(Z*A \cdot t^2 + Z*B \cdot t\right) \cdot negParens(E) =
X*E * B + C*C + Z*B*E \cdot t + D*C.
We must compute the (elementary) integral \displaystyle \int_\gamma K d\gamma =
\int_0^{L} K(\gamma(t)) \cdot \gamma'(t) dt =
\int_0^{L} \left(X*E * B + C*C + Z*B*E \cdot t + D*C \right ) dt
.
This is \displaystyle
\int_0^{L} \left(X*E * B + C*C + Z*B*E \cdot t + D*C \right ) dt =
\left( fractionReduce(X*E * B + C*C + Z*B*E,2) \cdot t^2 + D*C\cdot t \right)\bigg|_0^{L} =
fractionReduce((X*E * B + C*C + Z*B*E)*L*L + 2 * D*C*L,2)
.