Understanding ${\sigma}$ in $\oint F*dr=_{\sigma}\iint(curlF)nds$

[nhrock3]

$$_{c}\oint F*dr=_{\sigma}\iint(curlF)nds$$
F(x,y,z)=(2z)i+(3x)j+(5y)k
$$_{\sigma}$$ is a part of a paraboloid $$z=4-x^2-y^2$$ where z>=0
on the x-y plane our paraboloid is 4=x^2+y^2
and the parametric view of it is:
x=2cost y=2sint z=0

so we get
$$_{c}\oint F*dr= (2z)dx+(3x)dy+(5y)dz$$

i can't understand the next step$$<br /> _{c}\oint F*dr= (2z)dx+(3x)dy+(5y)dz=\intop_{0}^{2\pi}[0+(6cost)(2cost)+0]dt$$
why??

 

[tiny-tim]

hi nhrock3!

_{c}∫ F*dr=_{\sigma}∫(curlF)nds

F(x,y,z)=(2z)i+(3x)j+(5y)k

{\sigma} is a part of a paraboloid z=4-x^2-y^2 where z>=0
on the x-y plane our paraboloid is 4=x^2+y^2
and the parametric view of it is:
x=2cost y=2sint z=0

so we get

_{c}∫ F*dr= (2z)dx+(3x)dy+(5y)dz


i can't understand the next step$$<br /> _{c}∫ F*dr= (2z)dx+(3x)dy+(5y)dz=∫_{0}^{2\pi}[0+(6cost)(2cost)+0]dt<br /> <br /> why??$$

$$<br /> it comes from x=2cost y=2sint z=0 …<br /> <br /> (and so dy = 2cost dt) <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />$$