Answer
$\mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = \frac{\pi }{8}\sqrt 2 - \frac{1}{2}\sqrt 2 + \frac{\pi }{4} + \frac{1}{2}\ln 2$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y,z} \right) = \left( {z,{x^2},y} \right)$ and the path ${\bf{r}} = \left( {x,y,z} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z$
The path is given by ${\bf{r}}\left( t \right) = \left( {\cos t,\tan t,t} \right)$ for $0 \le t \le \frac{\pi }{4}$. So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {t,{{\cos }^2}t,\tan t} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - \sin t,{{\sec }^2}t,1} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = \mathop \smallint \limits_0^{\pi /4} {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = \mathop \smallint \limits_0^{\pi /4} \left( {t,{{\cos }^2}t,\tan t} \right)\cdot\left( { - \sin t,{{\sec }^2}t,1} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^{\pi /4} \left( { - t\sin t + 1 + \tan t} \right){\rm{d}}t$
(1) ${\ \ \ \ \ }$ $\mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = - \mathop \smallint \limits_0^{\pi /4} t\sin t{\rm{d}}t + \mathop \smallint \limits_0^{\pi /4} {\rm{d}}t + \mathop \smallint \limits_0^{\pi /4} \tan t{\rm{d}}t$
1. Evaluate $ - \mathop \smallint \limits_0^{\pi /4} t\sin t{\rm{d}}t$
Write $u = t$, $dv = \sin tdt$. So, $du = dt$, $v = - \cos t$.
Using Integration by Parts Formula (Section 8.1):
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
we get
$ - \mathop \smallint \limits_0^{\pi /4} t\sin t{\rm{d}}t = - \left( { - t\cos t|_0^{\pi /4} + \mathop \smallint \limits_0^{\pi /4} \cos t{\rm{d}}t} \right)$
$ = - \left( { - \frac{\pi }{4}\cdot\frac{1}{2}\sqrt 2 + \sin t|_0^{\pi /4}} \right)$
$ = \frac{\pi }{8}\sqrt 2 - \frac{1}{2}\sqrt 2 $
2. Using the result in Example 6 of Section 8.2, we obtain
$\mathop \smallint \limits_0^{\pi /4} \tan x{\rm{d}}x = \left( {\ln \left| {\sec x} \right|} \right)|_0^{\pi /4}$
$ = \ln \left( {\frac{1}{2}\sqrt 2 } \right) = \ln \sqrt 2 - \ln 2$
$ = \frac{1}{2}\ln 2$
Substituting these results back in equation (1) gives
$\mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = - \mathop \smallint \limits_0^{\pi /4} t\sin t{\rm{d}}t + \mathop \smallint \limits_0^{\pi /4} {\rm{d}}t + \mathop \smallint \limits_0^{\pi /4} \tan t{\rm{d}}t$
$ = \frac{\pi }{8}\sqrt 2 - \frac{1}{2}\sqrt 2 + \frac{\pi }{4} + \frac{1}{2}\ln 2$
So, $\mathop \smallint \limits_C^{} z{\rm{d}}x + {x^2}{\rm{d}}y + y{\rm{d}}z = \frac{\pi }{8}\sqrt 2 - \frac{1}{2}\sqrt 2 + \frac{\pi }{4} + \frac{1}{2}\ln 2$.
