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Please help me calculate integral on surface.

$$\iint\!(2-y)\,dS $$ $$ 0\leq z\leq 1;\, y =1 $$ I can't understand what should I do with '$z$' coordinate. I assume I should do something with it, not to leave blank. Maybe like this? $$= \iint\!dy dz = 1?$$

Stahl
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SevenDays
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  • Hint: with the integral of a single variable--that is, $y=f(x)$, do you do anything with the $y$, or just "leave it blank"? – rurouniwallace Jun 24 '13 at 17:01
  • What is the integral of the function within a constant $y$? That is: What is the integral of that function from $y = 1$ to $y = 1$? – NasuSama Jun 24 '13 at 17:02
  • I have only this input data. So I think $$=\int\int (2-1)*\sqrt(1+d(y)^2) ..= $$ don't know what to do with z-coordinate I get this integral from my home-work. @ZettaSuro , I replaced y with it value. – SevenDays Jun 24 '13 at 17:07
  • Another question. $2-y$ doesn't tell us anything about the surface. Is the surface $x=2-y$ or $z=2-y$? I'm assuming it's $x=2-y$ but I just want to be sure. – rurouniwallace Jun 24 '13 at 17:30
  • So I see you facing the same problem as me. As for me, I found that maybe the surface is a line? But can't figure out how to use z-coordinate in this equation. Maybe you can suggest me a best solution for this problem so I can write in my work? – SevenDays Jun 24 '13 at 17:39
  • It is absolutely unclear what your surface is. – Christian Blatter Jun 24 '13 at 17:54
  • Sorry, but this is an example from a book... – SevenDays Jun 24 '13 at 18:06
  • Here is the method. By the way, consider up voting the answer if benfit from it. – Mhenni Benghorbal Jun 25 '13 at 23:04

1 Answers1

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Most likely there is part of a surface $S$ which is bounded between $0\leq z\leq 1$ and $y=1$.

If the integral is on the surface $y=1$ and bounded between $0\leq z\leq 1$, then this surface is an infinite strip parallel to the $x$-axis, there is no way the integral is $1$ unless you have a bound on $x$ as well.

If it is the former case, what you wanna do is to look for a parametrization of $S$: $$ S: \mathbf{\Phi}(u,v) = \big(x(u,v),y(u,v),z(u,v)\big), $$ for $u\in [a,b]$, $v\in [c,d]$, then $$ \iint_S (2-y)\,dS = \int^d_c\int^b_a \big(2-y(u,v)\big)\,\left|\frac{\partial \mathbf{\Phi}}{\partial u}\times \frac{\partial \mathbf{\Phi}}{\partial v}\right|\,dudv. $$

EDIT: for the $z = 1$ case, you certain have: $$ \iint_S (2-z)\,dS = \int^{1}_0\color{red}{\int^{b}_{a}} (2-z)\sqrt{1 + (z_x)^2 + (z_y)^2} \,\color{red}{dx} dy = \color{red}{\int^{b}_{a}dx}\int^{1}_0(2-1)dy. $$ Like I said in the original answer, you have to have a bound on $x$, otherwise the surface is infinite.

Shuhao Cao
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  • As my teacher says for the another example, I think maybe in example below we can do the same $$z=1, ds = \sqrt{(1+(z'_x)^{2}+(z'_y)^{2}} =1 $$, So The answer will be $$ \int_0^1 (2-1) dy = 1$$ – SevenDays Jun 25 '13 at 20:57
  • @wsevendays Please see my edit. – Shuhao Cao Jun 25 '13 at 22:17
  • yes, you are right, I've missed x coordinate. So I think this example is incorrectly writed. – SevenDays Jun 25 '13 at 22:47