Q. 14 - Calculus

Question

Write ${\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r$ explicitly as an integral of t, where $\mathbf{F}(x,y,z)=(zy-x, xz-y, xy-z)$ and $\mathbf{r}\left(t\right)=(t, \cos t, \sin t)$ for $a \le t \le b$.

Step-by-Step Solution

Answer

Hence, ${\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r={\int }_a^b\sin t\cos t-t-t\cos 2tdt$

Step 1. Given Information

Write ${\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r$ explicitly as an integral of t, where $\mathbf{F}(x,y,z)=(zy-x, xz-y, xy-z)$ and $\mathbf{r}\left(t\right)=(t, \cos t, \sin t)$ for $a \le t \le b$.

Step 2. The curve parameterize by r ( t ) = ( t ,   cos t ,   sin t )   at   a ≤ t ≤ b

Thus $\mathbf{r}\mathbf{\text{'}}\left(t\right)=(1, -\sin t, \cos t)$. So,

$$\begin{gathered} {\int }_{C}f(x,y,z)\mathrm{dr}={\int }_a^{b}f\left(r\right(t)·\mathbf{r}\text{'}(t)dt \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{dr}={\int }_a^b(\sin t\cos t-t,t\sin t-\cos t,t\cos t-\sin t\left)\right(1,-\sin t,\cos t)dt \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r={\int }_a^{b}1(\sin t\cos t-t)-\sin t(t\sin t-\cos t)+\cos t(t\cos t-\sin t)dt \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r={\int }_a^b\sin t\cos t-t-t{\sin }^{2}t+\sin t\cos t+t{\cos }^{2}t-\sin t\cos tdt \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r={\int }_a^b\sin t\cos t-t+t({\cos }^{2}t-{\sin }^{2}t)dt \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}r={\int }_a^b\sin t\cos t-t-t\cos 2tdt \end{gathered}$$