Oct 1, 2010 #1 ƒ(x) Messages 327 Reaction score 0 Convert x+y = 1 to parametric equations. I know that z = v, but after that I'm stuck.
Oct 1, 2010 #2 Char. Limit Gold Member Messages 1,222 Reaction score 23 Wait, z=v? What's that got to do with anything? A simple way to do this is let x=t.
Oct 1, 2010 #3 ƒ(x) Messages 327 Reaction score 0 I need to convert it into three parametric equations for x, y, and z
Oct 1, 2010 #4 ZioX Messages 370 Reaction score 0 There are lots of parametrizations. Which one you would want to use depends on what you want to do with it.
There are lots of parametrizations. Which one you would want to use depends on what you want to do with it.
Oct 1, 2010 #5 ƒ(x) Messages 327 Reaction score 0 It's in the section of my book dealing with finding surface area by using double integrals.
Oct 1, 2010 #6 ƒ(x) Messages 327 Reaction score 0 its the infinite amount of parametrizations that throws me
Oct 1, 2010 #7 ZioX Messages 370 Reaction score 0 You typically pick one that cancels out factors in your integration.
Oct 1, 2010 #8 ƒ(x) Messages 327 Reaction score 0 ok, but my book wanted me to do it a certain way. like when the function was x^2 + y^2 = 1 z = V x = cosU y = sinU and then for another one they wanted z = V x = sinUcosV y = sinUsinV Last edited: Oct 1, 2010
ok, but my book wanted me to do it a certain way. like when the function was x^2 + y^2 = 1 z = V x = cosU y = sinU and then for another one they wanted z = V x = sinUcosV y = sinUsinV
Oct 1, 2010 #9 ZioX Messages 370 Reaction score 0 Right, because they come out nice. The point is that the parametization doesn't matter. The double integral will always evaluate to the same value. (t,1-t,z) will probably work fine for you.
Right, because they come out nice. The point is that the parametization doesn't matter. The double integral will always evaluate to the same value. (t,1-t,z) will probably work fine for you.