@ganeshie8 , trying to deal with a Surface Integral. I'm not sure what I'm doing wrong here, posted below momentarily.
Part of it, near the end; reposting the rest of it in a second. http://i.imgur.com/5nUli9o.png
\[G(x,y,z)=x; \]Integrated over the surface \[y = x^2, \ \ \ 0 \leq x \leq 2, \ \ \ 0 \leq z \leq 3\]
I parametrized the surface as such: \[u = x, \ \ \ v = z; \ \ \ r(u,v) = u, u^2, v\]
\[r_{u},r_{v}=(1,2u,0), \ \ \ (0,0,1)\]
Cross product of the two vectors yields \[2u \ i - j\]
Alright, one sec, lol, forgot what I did after this. One moment.
Oh yeah, parametrized the vector field, \[G(x,y,z)=G(r(u,v))=u\]
Dot product between the two vectors
(Sorry, was posting on another thread, back)
Dot product just yields 2u^2?
why not simply use the formula \[ \hat{n}dS = \langle -f_x, 1, -f_z\rangle dxdz\] ?
This is the book answer, have no freaking clue how they got there while I got somewhere very far away: http://i.imgur.com/soIJgWi.png
I don't know what that formula is, the formulae I'm supposed to be using are http://i.imgur.com/CbQYLfw.png I'm trying to use the first formula
But yeah, I have no idea what the formula is/where that came from, what you're using, I'm just trying to use the first formula in that picture to solve the problem, and I'm not sure what I'm doing wrong with it.
GOT IT, I think
I think your textbook is using the last formula
just change the area element to `dxdz`
\[\iint_R G(x,f(x,z),z )~ \sqrt{f_x^2+f_z^2+1} ~dxdz\]
Then just evaluating it
I got there using the first one
Alright, since I figured this one out, I'm going to take a shot at another problem using a different technique, or altogether a different problem. I'm running out of time.
Man, all of my messages are getting jumbled up, OpenStudy isn't playing well today
But yeah, I figured it out using that technique, going to move on to a different problem for now since I'm short on time.
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