$[$Sec$.13.$C$]$ This problem will provide some practice using different notations when computing vector line integrals. Consider the curve $C=(sint,e^{3t},t^4),0t1$ and the vector field $F=+y^{2}j+z^{3}k.$

$($a$)$ Compute ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}F*dr$ using the following steps.

i$.$ Write the curve in terms of position vectors, $r(t).$

ii$.$ Find $r^{\text{'}}(t).$

iii. Evaluate $F$ on the curve, that is$,$ find $F(r(t)).$

iv$.$ Compute $F(r(t))*r^{\text{'}}(t).$

v$.$ Set up$,$ but do not evaluate ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}F*dr$ as ${\displaystyle \underset{0}{\overset{1}{}}}F(r(t))*r^{\text{'}}(t)dt$

$($b$)$ Compute ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}xdx+y^{2}dy+z^{3}dz$ using the following steps.

i$.$ With $x=sint,y=e^{3t},z=t^4,$ compute $dx=x^{\text{'}}dt,dy=y^{\text{'}}dt,dz=z^{\text{'}}dt.$

ii$.$ Write $xdx+y^{2}dy+z^{3}dz={?}^{\text{'}}dt+y^2{y}^{\text{'}}dt+z^3{z}^{\text{'}}dt$ in terms of $t$ and $dt.$ It can be written in the form $(($:cdots

The answer is $1/2$sin$^2(1)+1/3($e$^9-1)+1/4=1/2-1/4$cos$2+1/3($e$^9-1)$ I$\text{'}$m good with the integral set up$,$ i dont understand how to get that specific answer though