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December 13th, 2016, 02:12 PM   #1
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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.
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December 13th, 2016, 03:49 PM   #2
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Please choose specific titles (and omit the swearing), thanks ...



I'v edited the title and deleted the duplicate thread.
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December 13th, 2016, 03:57 PM   #3
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thanks admin! solve this for me =D?
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December 13th, 2016, 04:07 PM   #4
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What have you tried?
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December 13th, 2016, 04:08 PM   #5
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i have no clue how to even start it... =/
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December 13th, 2016, 04:09 PM   #6
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i know that u guys are pro at math xD gimme just this one
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December 13th, 2016, 04:26 PM   #7
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Originally Posted by aneka View Post
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?
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December 13th, 2016, 05:14 PM   #8
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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.
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Last edited by skipjack; December 13th, 2016 at 07:11 PM. Reason: to change "tex" to "math"
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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!!

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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 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.
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Last edited by skipjack; December 13th, 2016 at 07:10 PM.
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December 13th, 2016, 06:30 PM   #10
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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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