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Compute $\int\int_\limits{\gamma}\vec{F}\cdot\bar{n} dS$, where

$\vec{F}=[x^{2}yz+xe^{z}]\bar{i}+[x^{2}+y(1-e^{z})]\bar{j}+[2+x^{3}-xyz^{2}]\bar{k}$

$\gamma =x^{2}+y^{2}=(z-1)^{2} ,0\leq z\leq 1$

Am not interested only in solving this particular example but in understanding how to work these kind of problems in general so that I can do the rest of problems alone. I need detailed answer , the more detailed the better .

Thanks in advance.

vidyarthi
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Kasmir Khaan
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1 Answers1

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Here is the general methodology for these problems:

  1. Sketch the surface if it is not too complicated: here it is the part of the cone shown below with $0\le z \le 1$:

enter image description here

  1. Find a parametrization of the surface on which you are integrating: here is one possibility: \begin{cases} x=x\\ y=y\quad\quad\quad\quad\quad\quad\quad x,y \; |\; 0 \le x^2+y^2 \le 1 \\ z=\sqrt{x^2+y^2}+1 \end{cases}

By experience, it is often a good idea to use $x$ and $y$ as parameters, as step 3. is always easy this way. It might be tempting to use parameters $r$ and $\theta$, but this may complicate the next step.

  1. Compute $\vec{r}_x \times \vec{r}_y$: $$ \vec{r}_x \times \vec{r}_y = (1,0,\frac{x}{\sqrt{x^2+y^2}}) \times (0,1,\frac{y}{\sqrt{x^2+y^2}}) = (-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1) $$
  2. Make sure the orientation is correct (it depends on the convention you are using). Here, the third component of $\vec{r}_x \times \vec{r}_y$ is positive, so I will assume the orientation is correct.

  3. You are ready to integrate: $$ \iint_{\gamma}\vec{F}\cdot d\vec{S} =\iint_{x,y| 0\le x^2+y^2 \le 1}\vec{F}(x,y)\cdot \vec{r}_x \times \vec{r}_y\; dx dy = \iint_{x,y| 0\le x^2+y^2 \le 1}f(x,y)\; dx dy $$

    1. Perform a change of variables if necessary: $$ \iint_{x,y| 0\le x^2+y^2 \le 1}f(x,y)\; dx dy = \int_{r=0}^1\int_{\theta=0}^{2\pi}f(r,\theta) \; r dr d\theta $$

    2. If possible, check your answer with a software, or more interestingly, by comparing it with another method such as the divergence theorem here. As pointed out by @Mattos, the problem is much simpler using the divergence theorem. I suspect your teacher wants you to try it the hard way to fully appreciate the divergence theorem next week :)

Kuifje
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  • Thanks alot very elegant and clear answer ! :) Can you please explain the orientation concept , you could have taking cross product of y first than x and would be the negative , and what am calculating from mathematical point of view only , thanks alot again – Kasmir Khaan Nov 30 '16 at 16:33
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    Such an integral represents the flux of the vector field $\vec{F}$ through surface $\gamma$. For example, if $\vec{F}$ is the gradient of a scalar field representing the temperature in space, this flux represents the heat transfer through the surface $\gamma$ which could represent a window. There are many other physical interpretations possible. – Kuifje Nov 30 '16 at 16:40
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    Regarding the orientation concept, as you said it will only affect the sign of the final answer. This concept is probably detailed in your course. Typically, for a surface $z=f(x,y)$, you will want the third component to be positive. For a surface $y=f(x,z)$ you will want the second component to be positive. And for a surface $x=f(y,z)$, you will want the first component to be positive. The orientation will be more crucial when you will dive into the divergence theorem. – Kuifje Nov 30 '16 at 16:43