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

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SUMMARY

The discussion focuses on the application of Stokes' Theorem, specifically the equation \(\oint F \cdot dr =_{\sigma} \iint (curl F) \, nds\), using the vector field \(F(x,y,z) = (2z)i + (3x)j + (5y)k\). The surface \(\sigma\) is defined as a portion of the paraboloid \(z = 4 - x^2 - y^2\) for \(z \geq 0\). The parametric representation of the curve is given by \(x = 2\cos(t)\), \(y = 2\sin(t)\), and \(z = 0\), leading to the integral \(\oint F \cdot dr = \int_{0}^{2\pi} [0 + (6\cos(t))(2\cos(t)) + 0] dt\). The confusion arises regarding the transition to the integral form, which is clarified by the relationship between the parametric equations and their derivatives.

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nhrock3
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<br /> _{c}\oint F*dr=_{\sigma}\iint(curlF)nds<br />
F(x,y,z)=(2z)i+(3x)j+(5y)k
<br /> _{\sigma}<br /> 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
<br /> _{c}\oint F*dr= (2z)dx+(3x)dy+(5y)dz<br />

i can't understand the next step<br /> <br /> _{c}\oint F*dr= (2z)dx+(3x)dy+(5y)dz=\intop_{0}^{2\pi}[0+(6cost)(2cost)+0]dt<br />
why??
 
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hi nhrock3! :smile:
nhrock3 said:
_{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 /> <br /> _{c}∫ F*dr= (2z)dx+(3x)dy+(5y)dz=∫_{0}^{2\pi}[0+(6cost)(2cost)+0]dt<br /> <br /> why??
<br /> <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:" />
 

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