December 13th, 2016, 02:12 PM  #1 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0  Area of bounded surface
godness =D just impossible this, pls step by step cos i need to understand i really appreciate Last edited by greg1313; December 13th, 2016 at 03:48 PM. 
December 13th, 2016, 03:49 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,949 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
Please choose specific titles (and omit the swearing), thanks ... I'v edited the title and deleted the duplicate thread. 
December 13th, 2016, 03:57 PM  #3 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
thanks admin! solve this for me =D?

December 13th, 2016, 04:07 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,949 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
What have you tried?

December 13th, 2016, 04:08 PM  #5 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
i have no clue how to even start it... =/

December 13th, 2016, 04:09 PM  #6 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
i know that u guys are pro at math xD gimme just this one

December 13th, 2016, 04:26 PM  #7 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,829 Thanks: 648 Math Focus: Yet to find out.  
December 13th, 2016, 05:14 PM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Have you at least taken a Calculus III class? Here's how I would start this. Write the surface as the vector equation $\displaystyle f(u,v)= <u \sin(v), u \cos(v), v>$. The derivatives with respect to u and v are $\displaystyle f_u= <\sin(v), \cos(v), 0>$ and $\displaystyle f_v= <u \cos(v), u \sin(v), 1>$. Those derivative vectors are both tangent to the surface. The [b]cross product[b] is perpendicular to the surface: $\displaystyle \left\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k}\\ u \cos(v) & u \sin(v) & 1 \\ \sin(v) & \cos(v) & 0 \end{array}\right= \cos(v)\vec{i}+ \sin(v)\vec{j}+ u\vec{k}$. The surface area is the length of that vector integrated over the u, v limits. Last edited by skipjack; December 13th, 2016 at 07:11 PM. Reason: to change "tex" to "math" 
December 13th, 2016, 05:32 PM  #9  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff. 
Country Boy do you ever look at your work after you've posted it? MATH tags!! Dan Quote:
Last edited by skipjack; December 13th, 2016 at 07:10 PM.  
December 13th, 2016, 06:30 PM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,949 Thanks: 1141 Math Focus: Elementary mathematics and beyond  

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