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thepatient
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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.
Evaluate the surface integral [itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex], 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.
[itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex] = [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex], where
d[itex]\sigma[/itex] = |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|dudv, and r is a vector parametricized for the surface.
n = outward unit normal vector from the surface.
First, I parametricized the surface as a vector in terms of u and v:
x = u
y = v
z = 2 -3x - 1.5y
[itex]\overline{r}[/itex](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[itex]\overline{r}[/itex]/du = <1, 0, -3>
d[itex]\overline{r}[/itex]/dv = <0,1,-1.5>
Calculated |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|:
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:
[itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex] R=[(u,v): 0<u<1, 0<v<2-2u]
[itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex] 1/7 (6u^2 +3uv + 4 - 6u - 3v) *(3.5) *dvdu =
1/2 [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]6u^2 + 3uv -6u - 3v + 4 dv du =
1/2 [itex]\int[/itex]6u^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 [itex]\int[/itex]24u^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.
Homework Statement
Evaluate the surface integral [itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex], 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
[itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex] = [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex], where
d[itex]\sigma[/itex] = |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|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
[itex]\overline{r}[/itex](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[itex]\overline{r}[/itex]/du = <1, 0, -3>
d[itex]\overline{r}[/itex]/dv = <0,1,-1.5>
Calculated |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|:
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:
[itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex] R=[(u,v): 0<u<1, 0<v<2-2u]
[itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex] 1/7 (6u^2 +3uv + 4 - 6u - 3v) *(3.5) *dvdu =
1/2 [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]6u^2 + 3uv -6u - 3v + 4 dv du =
1/2 [itex]\int[/itex]6u^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 [itex]\int[/itex]24u^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.
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