Recent content by KnowledgeisPo

That could be because you don't actually need to do it in the xy plane. And my integral was indeed defined from 0>z>1-y, and 0>y>1. If I was substituting z in the integral with z=1-x-y, then yes, I would want to integrate over the xy, and yes, if you wanted to point out a conceptual typo, you...

No joke. That's pretty much the way I explained it in the post above... Much appreciated if you were doing it while I was explaining the solution I got. I don't think I mentioned the cross product of the r_x and r_y vectors, but yeah, that's how you get the segment to integrate with the function...

ζζs yz ds S is the part of the plane x+y+z=1 that lies in the first Octant Homework Equations ζζs f(x,y,z) dS = ζζd f(x,y,g(x,y)) sqrt(1+(df/dx)²+(df/dy)²) dA solution sqrt(1+(df/dx)²+(df/dy)²)-- df/dx=-1, df/dy=-1 therefor sqrt(1+ (-1)² + (-1)²)= sqrt(3) ζζs f(x,y,z) dS=ζζs yz dS=ζζd...

The region on the xz would be x=1-z, the region on the xy would be y=1-x, over the yz would be z=1-y. Since the surface area with equation, x+y+z=1, is a triangle flush against the xz and yz planes, I set my dA to be the region on the XY plane, in which x and y go from 0 to 1. Every time I set...

Homework Statement ζζs yz ds S is the part of the plane x+y+z=1 that lies in the first OctantHomework Equations ζζs f(x,y,z) dS = ζζd f(x,y,g(x,y)) sqrt(1+(df/dx)²+(df/dy)²) dAThe Attempt at a Solution turned dS into dA, and attempted many different regions. Always got an area that was 4-6...

<2u+3v, 1+5u-v, 2+u+v> P(7,10,4) Q(5,22,5) P: 7=2u+3v, 10=1+5u-v, *4=2+u+v* 2=u+v, so u=2-v 7=2(2-v)+3v 7=4-2v+3v 3 =v,, 7=2u+9, u=-1 10=?=1+5(-1)-3,, 10=/=-7 Q: 5=2u+3v, 22=1+5u-v, *5=2+u+v* 3=u+v...

Yes. As I said a few posts, I have solved it. I was aware that multivariable parametric surfaces are surfaces. However, was I supposed to not say what the book said? Thank you again for the assistance.

Yes, this is a little bit why I didn't understand the book. I improvised on terminology a little bit when I said partial vectors, then continued to allocate them to I,J, and K components of what would be a function on a higher integration.

Solved Thank you for your time, and sorry for any inconvenience, as I solved it myself. I set the partial vectors equal to the respective points, and used the simple Z partial to put v in terms of u. Upon doing that, I substituted that in for the X partial to get u and v coordinates at the...

I see what you are saying. Keep in mind this is Parametric Surfaces and Their Areas. That was all the given information, and that is how the wording is transcribed in the book. Perhaps P and Q corresponds to the vector itself, and i and j of the vector are to correspond to P and Q respectively...

Homework Statement Determine whether the points P(7,10,4) and Q(5,22,5) lie on the given line: Homework Equations r(u,v)=<2u+3v, 1+5u-v, 2+u+v> The Attempt at a Solution x=x(u,v)=2u+3v, y=1+5u-v, z=2+u+v x+y+z=8u+3v+3