Calculating flux using surface integrals.

[thepatient]

This isn't homework. I've been restudying vector calculus from the beginning to end on my free time and got stuck on this problem. I am not sure what I'm doing wrong, but it may be a calculation error since it has so much calculation involved.

Homework Statement

Evaluate the surface integral \int\int_{\Sigma}f*d\sigma, where f(x, y, z) = x2i + xyj + zk and is the
part of the plane 6x + 3y + 2z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit
normal n pointing in the positive z direction.

Homework Equations

\int\int_{\Sigma}f*d\sigma = \int\int_{R}f*n*d\sigma, where
d\sigma = |dr/du X dr/dv|dudv, and r is a vector parametricized for the surface.

n = outward unit normal vector from the surface.

The Attempt at a Solution

First, I parametricized the surface as a vector in terms of u and v:
x = u
y = v
z = 2 -3x - 1.5y
\overline{r}(u,v) = <u, v, 2-3u - 1.5v>
R is the region on the u, v plane under the line v = 2 -2u, u≥0, v≥0
Took partial derivatives in terms of u and v:
d\overline{r}/du = <1, 0, -3>
d\overline{r}/dv = <0,1,-1.5>

Calculated |dr/du X dr/dv|:
i j k
1 0 -3
0 1 -1.5 =
|3i + 1.5j + k| = (9 + 2.25 + 1)^1/2 = 3.5

Calculated n:
Since the plane is 6x + 3y + 2z = 6, perpendicular to the plane would be the vector v = <6, 3, 2>, so n would be v/(|v|), the vector divided by its magnitude.
n = <6, 3, 2>/|7| = <6/7, 3/7, 2/7>

Calculating f(u,v) * n:
f(u,v) = <(u^2), (uv), (2 -3u - 1.5v)>* <6/7, 3/7, 2/7> =
1/7 (6u^2 +3uv + 4 - 6u - 3v)

Plugging into equation:
\int\int_{R}f*n*d\sigma R=[(u,v): 0<u<1, 0<v<2-2u]

\int\int_{R} 1/7 (6u^2 +3uv + 4 - 6u - 3v) *(3.5) *dvdu =

1/2 \int\int_{R}6u^2 + 3uv -6u - 3v + 4 dv du =
1/2 \int6u^2v + 3uv^2 -3v^2/2 + 4v (evaluating v from 0 to 2-2u) dv

Using wolfram alpha to evaluate (got tired of doing it by hand like 6 times XD) :
1/2 \int24u^3 + 30u^2 + 8u + 2 du from u = 0..1
1/2 * (24u^4/4 + 30u^3/3 + 8u^2/2 + 2u) from 0 to 1
1/2 (6 + 10+4+2) = 11

The answer should be 15/4 though. I'm not sure what I did wrong, maybe I missed something. Can anyone see what I did wrong in this problem? Thanks much.

 

[vela]

I don't know if it's just a typo, but you dropped the term -6u from the integrand on the step where you simplified 3.5/7 to 1/2 and reordered the terms.

EDIT: I get 7/4 as well.

 

[thepatient]

Oops. I redid it and got 13/2 with the -6u term. It came out to 13/2 though. :(

 

[thepatient]

If I remember correctly, since the surface Σ is closed, the divergence theorem applies right?

∫∫Rf*n*dσ = ∫∫∫div(f) dV

x is from 0 to one
y is from 0 to the line -2x + 2
z is from 0 to the plane 3 - 3x -3y/2

f(x, y, z) = x2i + xyj + zk
div(f) = d/dx i + d/dy j + d/dz k * x^2i + xyj + zk
= 2x + x + 1 = 3x+1

∫∫∫div(f) dV = ∫∫∫3x+1 dzdydx
∫∫3xz+z | z = 0..3-3x-3y/2 dy dx
∫∫3x(3-3x-3y) + 3-3x-3y -[ 0] dydx
∫∫9x -9x^2 - 9xy +3 -3x - 3y dy dx
∫∫6x - 9x^2 -9xy +3 -3y dy dx
∫6xy - 9x^2y - 9xy^2/2 + 3y - 3y^2/2 | y = 0..-2x+2 dx
∫6x(-2x+2) - 9x^2(-2x+2) -9x(-2x+2)^2/2 +3(-2x+2) -3(-2x+2)^2/2 dx
∫-12x^2 + 12x +18x^3-18x^2-(9x/2) (4x^2 -8x + 4) -6x + 6 -3(4x^2 - 8x + 4) dx=
∫-12x^2 + 12x + 18x^3 -18x^2 -36x^3/2 + 72x^2/2 -36x/2 + 6 -12x^2 +24x - 12 dx=
-12x^3/3 + 12x^2/2 + 18x^4/4 -18x^3/3 -36x^4/8 + 72x^3/6 - 36x^2/4 +6x -12x^3/3 +12x^2 -12x from x =0..1
-12/3 + 12/2 + 18/4 -18/3 -36/8 +12 -9 +6 -4 + 12-12 =1


:\ I'm frustrated with this problem.

 

[vela]

You don't have a closed surface. You're calculating the flux only on one plane. The volume integral of the divergence would equal the total flux through the plane and the other three sides of the tetrahedron.

 

[thepatient]

Hmm.. But the problem states "...part of the plane 6x + 3y + 2z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n pointing in the positive z direction." That doesn't mean that the flux is the net flux of the closed surface, but the flux of only that portion of the plane?

 

[vela]

First, I parametricized the surface as a vector in terms of u and v:
x = u
y = v
z = 2 -3x - 1.5y

You solved for z incorrectly. It should be z = 3 - 3x - (3/2)y = 3 - 3u - (3/2)v.

\overline{r}(u,v) = <u, v, 2-3u - 1.5v>
R is the region on the u, v plane under the line v = 2 -2u, u≥0, v≥0
Took partial derivatives in terms of u and v:
d\overline{r}/du = <1, 0, -3>
d\overline{r}/dv = <0,1,-1.5>

Calculated |dr/du X dr/dv|:
i j k
1 0 -3
0 1 -1.5 =
|3i + 1.5j + k| = (9 + 2.25 + 1)^1/2 = 3.5

Calculated n:
Since the plane is 6x + 3y + 2z = 6, perpendicular to the plane would be the vector v = <6, 3, 2>, so n would be v/(|v|), the vector divided by its magnitude.
n = <6, 3, 2>/|7| = <6/7, 3/7, 2/7>

All this is fine.

Calculating f(u,v) * n:
f(u,v) = <(u^2), (uv), (2 -3u - 1.5v)>* <6/7, 3/7, 2/7> =
1/7 (6u^2 +3uv + 4 - 6u - 3v)

You propagated the error from above, but your method is okay. With the correct expression for z, you would get
\frac{1}{2}\int_0^1\int_0^{2(1-u)} (6u^2+3uv+6-6u-3v)\,dv\,du = \frac{7}{4}

Plugging into equation:
\int\int_{R}f*n*d\sigma R=[(u,v): 0<u<1, 0<v<2-2u]

\int\int_{R} 1/7 (6u^2 +3uv + 4 - 6u - 3v) *(3.5) *dvdu =

1/2 \int\int_{R}6u^2 + 3uv -6u - 3v + 4 dv du =
1/2 \int6u^2v + 3uv^2 -3v^2/2 + 4v (evaluating v from 0 to 2-2u) dv

Using wolfram alpha to evaluate (got tired of doing it by hand like 6 times XD) :
1/2 \int24u^3 + 30u^2 + 8u + 2 du from u = 0..1
1/2 * (24u^4/4 + 30u^3/3 + 8u^2/2 + 2u) from 0 to 1
1/2 (6 + 10+4+2) = 11

The answer should be 15/4 though. I'm not sure what I did wrong, maybe I missed something. Can anyone see what I did wrong in this problem? Thanks much.

 

[vela]

Hmm.. But the problem states "...part of the plane 6x + 3y + 2z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n pointing in the positive z direction." That doesn't mean that the flux is the net flux of the closed surface, but the flux of only that portion of the plane?

Yes, I think so, otherwise it wouldn't specify which direction the normal points in.EDIT: It turns out you get the same answer, 7/4, because the vector field vanishes on the other three faces, so there's no flux through them.

 

[vela]

If I remember correctly, since the surface Σ is closed, the divergence theorem applies right?

∫∫Rf*n*dσ = ∫∫∫div(f) dV

x is from 0 to one
y is from 0 to the line -2x + 2
z is from 0 to the plane 3 - 3x -3y/2

f(x, y, z) = x2i + xyj + zk
div(f) = d/dx i + d/dy j + d/dz k * x^2i + xyj + zk
= 2x + x + 1 = 3x+1

∫∫∫div(f) dV = ∫∫∫3x+1 dzdydx
∫∫3xz+z | z = 0..3-3x-3y/2 dy dx
∫∫3x(3-3x-3y) + 3-3x-3y -[ 0] dydx

You dropped the factor of 1/2 on the y term.

I would pull the (3x+1) out front since it's a constant for the y and z integrations: \int_0^1 (3x+1)\int_0^{2(1-x)} \int_0^{3-3x-3/2 y}dz\,dy\,dxIt'll simplify the algebra a bit. After you integrate over z, you'll have\int_0^1 (3x+1)\int_0^{2(1-x)} (3-3x-3/2 y)\,dy\,dx = \int_0^1 3(3x+1)\int_0^{2(1-x)} [(1-x)-y/2]\,dy\,dx
If you keep things in terms of 1-x, I think the result of the y integral will be pretty simple. To do the x integration, you'll just have to bite the bullet and multiply it out, or try the substitution u=1-x.

∫∫9x -9x^2 - 9xy +3 -3x - 3y dy dx
∫∫6x - 9x^2 -9xy +3 -3y dy dx
∫6xy - 9x^2y - 9xy^2/2 + 3y - 3y^2/2 | y = 0..-2x+2 dx
∫6x(-2x+2) - 9x^2(-2x+2) -9x(-2x+2)^2/2 +3(-2x+2) -3(-2x+2)^2/2 dx
∫-12x^2 + 12x +18x^3-18x^2-(9x/2) (4x^2 -8x + 4) -6x + 6 -3(4x^2 - 8x + 4) dx=
∫-12x^2 + 12x + 18x^3 -18x^2 -36x^3/2 + 72x^2/2 -36x/2 + 6 -12x^2 +24x - 12 dx=
-12x^3/3 + 12x^2/2 + 18x^4/4 -18x^3/3 -36x^4/8 + 72x^3/6 - 36x^2/4 +6x -12x^3/3 +12x^2 -12x from x =0..1
-12/3 + 12/2 + 18/4 -18/3 -36/8 +12 -9 +6 -4 + 12-12 =1:\ I'm frustrated with this problem.

Either way, you should get 7/4 for the answer, so I wouldn't bother beating your head against the wall to try get 15/4. That answer is apparently wrong.

 

[thepatient]

You're right. I should just try the problems on my other calculus book. This book was a free online book which I think didn't have much revisement. Thanks much for your help. :]