Answer
a) $f(x,y,z)=x\sin y+y\cos z+K$
b) $1-\pi/2$
Work Step by Step
Let’s find a function $f$ such that $\nabla f=F$. We have:
$F(x,y,z)=$
$C: r(t)=<\sin t, t, z> 0<=t<=\pi/2$
First we see if the vector field is conservative
P or dxf= siny Q or dyf=xcosy+cosz R or dzf=-ysinz
dP/dy=cosy dQ/dx=cosy
=> the vector field is conservative
now we find the integral of dxf
$\int dxfdx =f(x,y,z)$
$\int sinydx =xsiny+f(y,z)$
$f(x,y,z)=xsiny+f(y,z)$
Now we find the derivative in respect to y of f(x,y,z) and set it equal to dyf
$d/dy(xsiny+f(y,z))=dyf$
$xcosy+f'(y,z)=xcosy+cosz$
$f'(y,z)=cosz$
$f(y,z)=sinz+h(z)$
Now we plug this into f(x,y,z)
$f(x,y,z)=xsiny+sinz+h(z)$
Now we find the derivative of z in f(x,y,z) and set it equal to dzf
$d/dz(xsiny+sinz+h(z)=dzf$
$cosz+h'(z)=-ysinz$
$h'(z)=-ysinz-cosz$
$h(z)=ycosz-sinz+K$
Now we plug in h(z) into our updated f(x,y,z)
$f(x,y,z)=xsiny+sinz+h(z)$ ->
$f(x,y,z)=xsiny+sinz+ycosz-sinz+K$
$f(x,y,z)=xsiny+ycosz+K$
Now we plug in our t values into the given curve;
$C: r(t)= 0<=t<=\pi/2$
$r(\pi/2)=(1,\pi/2,\pi)$
$r(0)=(0,0,0)$
Now we do FTC = f(b)-f(a)
b = $( 1,\pi/2,\pi)$
a =(0,0,0)
$f(1,\pi/2,\pi)-f(0,0,0)$
Plug this into our f(x,y,z)
$1sin\pi/2+\pi/2cos\pi-(0)=1-\frac{\pi}{2}$
