# Thread: help with area vectors

1. ## help with area vectors

ive done 15 pages of a calc III take home test, and this is the only thing im having trouble with.

so i seem to have taken poor notes when going over the concept of area vectors. i think my handwriting says that for flat surfaces, the area vector is the total area times the normal vector. is this true?

really, my problem is given a vector field f = ai + bj + ck where a,b, and c are constants,

4) let s be the disk of radius 2010 centered on the x-axis, in the plane x = -1, oriented away from the yz-plane. find the condition(s) on the constants a,b,and c such that the flux of f through s i positive or explain why no such condition exists.

5) let s be a triangle with vertices P (1,0,0), Q(0,2,0) and R(0,0,3) oriented away from the origin. same question.

6) let s be the unit sphere centered at (-4,29,-10) oriented outward. same question.

4) so if i know the area vector, then i can dot it with the vector field to find the flux. once they're crossed, it will be obvious what conditions will allow flux to be positive.

5) the same is true for this one, right? if i know the area vector, i can just dot it?

6) i have no idea since it's not a flat surface. the only thing we have learned is the divergence theorem, but this won't get me anywhere being that the derivatives of a, b, and c must be zero since they are constant.

any help would be greatly appreciated.

2. ok so what i have so far is this: (this is all assuming my whole spiel on are vectors is true)

4) the normal will be <-1,0,0> the area is pi(2010)^2

thus the area vector is <-pi(2010)^2,0,0>

the flux is then A dot V:

<-pi(2010)^2,0,0> dot <a,b,c>

= -a(pi)(2010)^2

this is positive for all negative, non-zero values of a.

5) pr cross pq is <-1,0,3> cross <-1,2,0>

which is <-6,-3,-2>

to make it point away from the origin as mandated, <6,3,2>

magnitude is 7, thus n hat is

<6/7,3/7,2/7>

area of triangle is .5 pr cross pq

= 7/2

therefore the area times n hat is

<3,3/2,1>

dotting with v, we get
3a + 3b/2 + c = flux

therefore, flux is positive for all values of a,b, and c such that

3a + 3b/2 + c >0

6)because s is closed, we can apply the divergence theorem.

however, this returns zero, as the derivative of constants is always zero.

therefore, no matter what values of a b and c are used, flux is always zero.

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edit

can someone just verify that all this makes sense and that i'm not misinterpreting the theorems?

3. Originally Posted by barriboy
ive done 15 pages of a calc III take home test, and this is the only thing im having trouble with.

so i seem to have taken poor notes when going over the concept of area vectors. i think my handwriting says that for flat surfaces, the area vector is the total area times the normal vector. is this true?
Yes, that is true. You may also have seen a theorem that says that if $\vec{u}$ and $\vec{v}$ are two vectors forming the adjacent sides of a parallelogram, then the area of the parallelogram is given by $|\vec{u}\times\vec{v}|$. Further the cross product is normal to the two vectors and so to the the plane the parallelogram is in. That is, the "area vector" is $\vec{u}\times\vec{v}$.

really, my problem is given a vector field f = ai + bj + ck where a,b, and c are constants,

4) let s be the disk of radius 2010 centered on the x-axis, in the plane x = -1, oriented away from the yz-plane. find the condition(s) on the constants a,b,and c such that the flux of f through s i positive or explain why no such condition exists.
The flux of a vector field through a surface is the integral of the dot product of the vector in the field with the "differential" area vector at each point. Since "area" is always positive, you don't need to worry about the area itself. That will be positive on if the dot product with the unit normal is positive. Here, the unit normal vector to the plane x= -1 is either $\vec{i}$ or $-\vec{i}$. Which of those is "oriented" away from the yz-plane? What is the dot product of f with that?

[quote]5) let s be a triangle with vertices P (1,0,0), Q(0,2,0) and R(0,0,3) oriented away from the origin. same question.
That gives the plane $\frac{x}{1}+ \frac{y}{2}+ \frac{z}{3}= 1$ (just check that each of those points satifies the equation). A normal vector is $\vec{i}+ \frac{1}{2}\vec{j}+ \frac{1}{3}\vec{k}$ which is "oriented away from the origin" because all of its components are positive. Divide by its length to get a unit vector, the take the dot product.

6) let s be the unit sphere centered at (-4,29,-10) oriented outward. same question.
The sphere is given by $(x+ 4)^2+ (y- 29)^2+ (z+ 10)^2= 1$.
You can think of that as a "level surface" for the function $f(x,y,z)= (x+ 4)^2+ (y- 29)^2+ (z+ 10)^2$ and you should recall that the gradient of a surface is always normal to it. The normal to that sphere at any point is $\nabla f= 2(x+ 4)\vec{i}+ 2(y- 29)\vec{j}+ 2(z+ 10)^2\vec{k}$. Again, take the dot product with the given function.

I'm not sure why you would use the divergence theorem. You are not asked what the flux is, only if it is positive.