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Math Help - surfance intergral question

  1. #1
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    surfance intergral question

    Hello

    I need help with a surface integral problem. I don't seem to understand how to approach it.

    My question is attached


    Thanks
    Attached Thumbnails Attached Thumbnails surfance intergral question-capture.gif  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    First step :

    Compute \vec{F}\cdot \vec{n} . Let's see what do you obtain.


    Fernando Revilla
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  3. #3
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    F.n=1-\frac{x}{2}-\frac{y}{2}

    thanks
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by cooltowns View Post
    F.n=1-\frac{x}{2}-\frac{y}{2}

    It is:

    \vec{F}\cdot \vec{n}=(x,0,-z)\cdot \dfrac{1}{\sqrt{6}}(1,1,2)= \dfrac{1}{\sqrt{6}}(x-2z)

    Now, use the formula that computes the surface integral by means of a double integral.


    Fernando Revilla
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  5. #5
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    I have always disliked the formula " \vec{F}\cdot\vec{n}dS" or simply " \vec{n}dS" and prefer d<br />
\vec{S} and \vec{F}\cdot d\vec{S} for the following reasons:

    To find the "differential of surface area", dS, for a surface given by parametric equations x= f(u,v), y= g(u,v), z= h(u,v) (or z= F(x,y) where you can just use "x" and "y" as parameters: x= x, y= y, z= F(x,y)), write the surface in "vector form" \vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}, take the derivatives with respect to the parameters, to get \vec{r}_u and \vec{r}_v, and, finally, take the cross product of those, \vec{r}_u\times \vec{r}_v. The resulting vector. called the "fundamental vector product" for the surface, will be perpendicular to the surface and its length, times du and dv, will be the "differential of surface area". But you see that using " \vec{n}dS, literally, means finding \vec{n}, the unit normal vector, which requires dividing that vector by its length while then multiplying by dS is just multiplying by that length again!

    A better way of thinking, in my opinion, is that you just integrate \vec{F}\cdot d\vec{S}, where the "vector differential of surface area" is just d\vec{S}= \vec{r}_u\times\vec{r}_v du dv.

    (Of course, the sign will depend on the order in which you multiply: \vec{r}_u\times\vec{r}_v= -\vec{r}_v\times\vec{r}_u. That is as it should be since you have a choice of orientation- which direction normal vector you take.)

    For this particular problem x+ y+ 2z= 2 or x= 2- y- 2z so we can take y and z (rather than x and y) as parameters: \vec{r}(y, z)= x\vec{i}+ y\vec{j}+ z\vec{k}= (2- y- 2z)\vec{i}+ y\vec{j}+ z\vec{k}. Then \vec{r}_y= -\vec{i}+ \vec{j} and \vec{r}_z= -2\vec{i}+ \vec{k}.

    The "fundamental vector product" for the surface, the cross product of those two vectors, is
    \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & 0 \\ -2 & 0 & 1\end{array}\right|= \vec{i}+ \vec{j}+ 2\vec{k}

    Now, I could calculate that the length of that is \sqrt{1+ 1+ 4}= \sqrt{6} so that \vec{n}= \frac{1}{\sqrt}6}\vec{i}+ \frac{1}{\sqrt}6}\vec{j}+ \frac{2}{\sqrt}6}\vec{k} and that dS= \sqrt{6}dydz but then, multiplying \vec{n}dS would just cancel the " \sqrt{6}"s so there was no need to calculate those in the first place! Just use d\vec{S}= \left(\vec{i}+ \vec{j}+2\vec{k}\right)dy dz.

    The question itself is ambiguous- there are two unit normal vectors at each point in opposite directions and whether you write the integral \int \vec{F}\cdot\vec{n}dS or \int \vec{F}\cdot d\vec{S} the sign will depend on which you choose. Since this plane cuts the axes in (2, 0, 0), (0, 2, 0), and (0, 0, 1), the normals could be distiguished as "up" (with positive z component) or "down" (with negative z component).

    Going around the boundary in the "positive orientation" ((2, 0, 0) to (0, 2, 0) to (0, 0, 1)) which is the most commonly used orientation, you would use the upward (z component positive) normal: d\vec{S}= \left(\vec{i}+ \vec{i}+2\vec{k}\right)dy dz.

    To find the limits of integration, since, here, we are integrating with respect to y and z, project the three vertices, (2, 0, 0), (0, 2, 0), and (0, 0, 1), onto the yz-plane: (y, z)= (0, 0), (2, 0), and (0, 2) and integrate over that triangle. If you choose to integrate as \int\int dydz you can see that z must go between 0 and 1. For each z, what will the conditions on y be to stay inside that triangle? What is the equation of the line between y= 2, z= 0 and y= 0, z= 1?
    Last edited by HallsofIvy; January 24th 2011 at 11:13 AM.
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  6. #6
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    the solution to this question, it simply said "this is the region
    bounded by the lines x = 2; y = 2 and x + y = 2", but it gives no explanation to where this came from, it then uses this information to define the limits of the double intergral.
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