Answer:
$4.35
Step-by-step explanation:
because 15% of 29 is 4.35
Answer:
27. 16/28. 48/29. 12/30. 33.3333333333
Step-by-step explanation:
Parameterize the surface (call it
) that has
as its boundary by
![\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(81-u^2)\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20s%28u%2Cv%29%3Du%5Ccos%20v%5C%2C%5Cvec%5Cimath%2Bu%5Csin%20v%5C%2C%5Cvec%5Cjmath%2B%2881-u%5E2%29%5C%2C%5Cvec%20k)
with
and
.
Take the normal vector to
to be
![\vec n=\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20n%3D%5Cdfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20u%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20v%7D%3D2u%5E2%5Ccos%20v%5C%2C%5Cvec%5Cimath%2B2u%5E2%5Csin%20v%5C%2C%5Cvec%5Cjmath%2Bu%5C%2C%5Cvec%20k)
Compute the curl of
. We have
![\nabla\times\vec F(x,y,z)=4y\,\vec\imath-4x\,\vec\jmath](https://tex.z-dn.net/?f=%5Cnabla%5Ctimes%5Cvec%20F%28x%2Cy%2Cz%29%3D4y%5C%2C%5Cvec%5Cimath-4x%5C%2C%5Cvec%5Cjmath)
![\implies\nabla\times\vec F(u,v)=4u\sin v\,\vec\imath-4u\cos v\,\vec\jmath](https://tex.z-dn.net/?f=%5Cimplies%5Cnabla%5Ctimes%5Cvec%20F%28u%2Cv%29%3D4u%5Csin%20v%5C%2C%5Cvec%5Cimath-4u%5Ccos%20v%5C%2C%5Cvec%5Cjmath)
Then by Stoke's theorem, the line integral is
![\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F(x,y,z))\cdot\mathrm d\vec S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r%3D%5Ciint_S%28%5Cnabla%5Ctimes%5Cvec%20F%28x%2Cy%2Cz%29%29%5Ccdot%5Cmathrm%20d%5Cvec%20S)
![=\displaystyle\iint_S(\nabla\times\vec F(u,v))\cdot\vec n\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Ciint_S%28%5Cnabla%5Ctimes%5Cvec%20F%28u%2Cv%29%29%5Ccdot%5Cvec%20n%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\displaystyle\int_0^{2\pi}\int_0^90\,\mathrm du\,\mathrm dv=\boxed{0}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E90%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B0%7D)
We can verify this result by computing the line integral directly. Parameterize
by
![\vec r(t)=9\cos t\,\vec\imath+9\sin t\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%28t%29%3D9%5Ccos%20t%5C%2C%5Cvec%5Cimath%2B9%5Csin%20t%5C%2C%5Cvec%5Cjmath)
with
. Then
![\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\vec F(t)\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cvec%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cvec%20r%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cvec%20F%28t%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%5Cvec%20r%7D%7B%5Cmathrm%20dt%7D%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^{2\pi}(9\cos t\,\vec\imath+9\sin t\,\vec\jmath+162\,\vec k)\cdot(-9\sin t\,\vec\imath+9\cos t\,\vec\jmath)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%289%5Ccos%20t%5C%2C%5Cvec%5Cimath%2B9%5Csin%20t%5C%2C%5Cvec%5Cjmath%2B162%5C%2C%5Cvec%20k%29%5Ccdot%28-9%5Csin%20t%5C%2C%5Cvec%5Cimath%2B9%5Ccos%20t%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^{2\pi}0\,\mathrm dt=\boxed{0}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D0%5C%2C%5Cmathrm%20dt%3D%5Cboxed%7B0%7D)
Answer:1689
Step-by-step explanation:
Delta Math
Answer:
What you said doesnt make sense
Step-by-step explanation:
"
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