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 May 26th, 2017, 08:57 AM #1 Newbie   Joined: Apr 2017 From: Brazil Posts: 5 Thanks: 0 Line Integral: intersection of sphere and cylinder Vectorial Field: F(x, y, z) = (y², z², x²) Sphere: x² + y² + z² = a² Cylinder: x² + y² = ax z<=0 ; a>0 How to parametrize the curve?? The exercise says it's clockwise when seen by the xy plan
 May 26th, 2017, 01:58 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I would start by using the standard parameterization for a sphere pf radius a (spherical coordinates): $x= a cos(\theta)sin(\phi)$ $y= a sin(\theta)sin(\phi)$ $z= a cos(\phi)$ Then $x^2+ y^2= a^2cos^2(\theta)sin^2(\phi)+ a^2sin^2(\theta)sin^2(\phi)= a^2 sin^2(\phi)$ and $ax= a^2 cos(\theta)sin(\phi)$ so that $sin(\phi)= cos(\theta)$. And, from that, $cos(\phi)= \sqrt{1- sin^2(\phi)}= \sqrt{1- cos^2(\theta)}= sin(\theta)$. Putting those together $x= a cos(\theta)sin(\phi)= a cos^2(\theta)$ $y= a sin(\theta)sin(\phi)= a sin(\theta)cos(\theta)$ $z= a cos(\phi)= a sin(\theta)$ Thanks from BadWolf Last edited by Country Boy; May 26th, 2017 at 02:42 PM.
May 27th, 2017, 01:27 PM   #3
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 Originally Posted by Country Boy I would start by using the standard parameterization for a sphere pf radius a (spherical coordinates): $x= a cos(\theta)sin(\phi)$ $y= a sin(\theta)sin(\phi)$ $z= a cos(\phi)$ Then $x^2+ y^2= a^2cos^2(\theta)sin^2(\phi)+ a^2sin^2(\theta)sin^2(\phi)= a^2 sin^2(\phi)$ and $ax= a^2 cos(\theta)sin(\phi)$ so that $sin(\phi)= cos(\theta)$. And, from that, $cos(\phi)= \sqrt{1- sin^2(\phi)}= \sqrt{1- cos^2(\theta)}= sin(\theta)$. Putting those together $x= a cos(\theta)sin(\phi)= a cos^2(\theta)$ $y= a sin(\theta)sin(\phi)= a sin(\theta)cos(\theta)$ $z= a cos(\phi)= a sin(\theta)$
Great! But.. The integral will be VERY laborious.. Any thoughts?

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line integral of x² y² z²=a²

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