My Math Forum Area of bounded surface

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 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: 7,968 Thanks: 1152 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. Thanks from topsquark and Joppy
 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: 7,968 Thanks: 1152 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
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Quote:
 Originally Posted by aneka i know that u guys are pro at math xD gimme just this one
No body is any help to you unless you let us know what you don't understand!

I dare say you won't be able to post your questions during a test . If you need a worked example, i'm sure your textbook has some?

 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 $\displaystyle f(u,v)=$. The derivatives with respect to u and v are $\displaystyle f_u= <\sin(v), \cos(v), 0>$ and $\displaystyle f_v=$. 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. Thanks from greg1313 Last edited by skipjack; December 13th, 2016 at 07:11 PM. Reason: to change "tex" to "math"
December 13th, 2016, 05:32 PM   #9
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Country Boy do you ever look at your work after you've posted it? MATH tags!!

-Dan

Quote:
 Originally Posted by Country Boy Have you at least taken a Calculus III class? Here's how I would start this. Write the surface as the vector equation $\displaystyle f(u,v)=$. The derivatives with respect to u and v are $\displaystyle f_u= <\sin(v), \cos(v), 0>$ and $\displaystyle f_v=$. Those derivative vectors are both tangent to the surface. The cross product 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 07:10 PM.

December 13th, 2016, 06:30 PM   #10
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Quote:
 Originally Posted by topsquark Country Boy do you ever look at your work after you've posted it? MATH tags!! -Dan
He's been reminded/edited/pm'd numerous times. Maybe we should call him "Tex"!

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