Graph the surface represented by the vector valued function

1. Nov 19, 2009

stevecallaway

1. The problem statement, all variables and given/known data
r(u,v)=2ucosv i + 2usinv j + u^4
0<=u<=1
0<=v<=2pi

2. Relevant equations
don't care about the graph as much as how the book got to the answer equation
z=(x^2+y^2)^2/16

3. The attempt at a solution[/b
cosu=x/2v sinu=y/2v cosu^2+sinu2=1 (x/2u)^2+(y/2u)^2=1 x^2+y^2=(4)^2
(x^2+y^2)=(4z)^2 (x^2+y^2)/16=z^2 z=sqrt((x^2+y^2)/16)
Why isn't the final answer z=sqrt...

Last edited: Nov 19, 2009
2. Nov 19, 2009

lanedance

x=2ucosv
y=2usinv
z=u^4

try substituting x & y into the equation
z=(x^2+y^2)^2/16
and see what you get

3. Nov 19, 2009

Dick

What is x^2+y^2 in terms of u?

4. Nov 19, 2009

stevecallaway

I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?

5. Nov 19, 2009

Dick

It's pretty garbled. But I'm with you through x^2+y^2=16*u^2, which is what I hope you meant to write. Then did you just put u=z? What did you do?

6. Nov 19, 2009

stevecallaway

Sorry about the garbled-ness. I thought I put more spaces in between the equations. Yes, I did substitute u=z and then I dividev both sides by 16 to get (x^2+y^2)/16=z^2. Then I took the sqrt of both sides which would give z=sqrt((x^2+y^2)/16) but the book gives an answer of z=(x^2+y^2)^2/16

7. Nov 19, 2009

Dick

But you can't put u=z! z=u^4. What does that make u^2?

8. Nov 19, 2009

stevecallaway

u^2=sqrt(z)...got it. Brilliant. Thanks.