- #1
WMDhamnekar
MHB
- 376
- 28
Evaluate the surface integral $\iint\limits_{\sum} f \cdot d\sigma $ where $ f(x,y,z) = x^2\hat{i} + xy\hat{j} + z\hat{k}$ and $\sum$ is the part of the plane 6x +3y +2z =6 with x ≥ 0, y ≥ 0,
z ≥ 0 , with the outward unit normal n pointing in the positive z direction.
My attempt to answer this question:
We need to parametrize the $\sum$. As we project $\sum$ onto xy-plane, it yields triangular region R = {(x,y): 0 ≤ x≤ 1, 0 ≤ y ≤ (2-2x) }. Thus, using (u,v) instead of (x,y), we see that,
x=u, y =v , z= 3-3u-3v/2 for 0 ≤ u ≤ 1, 0 ≤ v ≤ (2-2u) is the parametrization of $\sum$ over R (since z = 3-3x-3y/2 on $\sum$)
For (u,v) in R and for r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k = ui + vj + (3-3u-3v/2)k we have
$\displaystyle\left\vert\frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \displaystyle\right\vert= [1,0,-3] \times[ 0,1,-3/2] = [3, 3/2 ,1] \Rightarrow \left\vert \frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \right\vert= \frac72$
Thus, integrating over R using vertical slices $\iint\limits_{\sum} f\cdot d\sigma = \iint\limits_{\sum} f\cdot n d\sigma $
$ \iint\limits_{\sum} f\cdot d\sigma=\iint\limits_{R} (f(x(u,v), y(u,v), z(u,v))\cdot n )\left\vert \frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \right\vert dv du $
$\iint\limits_{\sum} f\cdot d\sigma = \int_0^1 \int_0^{2-2u} (\frac67 u^2 + \frac{3uv}{7} + \frac67 -\frac67 u -\frac37 v )\frac72 dv du $
$\iint\limits_{\sum} f\cdot d\sigma = \frac74 $
But the answer provided is $\frac{15}{4}.$How is that? Where are we wrong in the computation of answer?
z ≥ 0 , with the outward unit normal n pointing in the positive z direction.
My attempt to answer this question:
We need to parametrize the $\sum$. As we project $\sum$ onto xy-plane, it yields triangular region R = {(x,y): 0 ≤ x≤ 1, 0 ≤ y ≤ (2-2x) }. Thus, using (u,v) instead of (x,y), we see that,
x=u, y =v , z= 3-3u-3v/2 for 0 ≤ u ≤ 1, 0 ≤ v ≤ (2-2u) is the parametrization of $\sum$ over R (since z = 3-3x-3y/2 on $\sum$)
For (u,v) in R and for r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k = ui + vj + (3-3u-3v/2)k we have
$\displaystyle\left\vert\frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \displaystyle\right\vert= [1,0,-3] \times[ 0,1,-3/2] = [3, 3/2 ,1] \Rightarrow \left\vert \frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \right\vert= \frac72$
Thus, integrating over R using vertical slices $\iint\limits_{\sum} f\cdot d\sigma = \iint\limits_{\sum} f\cdot n d\sigma $
$ \iint\limits_{\sum} f\cdot d\sigma=\iint\limits_{R} (f(x(u,v), y(u,v), z(u,v))\cdot n )\left\vert \frac{\partial{r}}{\partial{u}} \times \frac{\partial{r}}{\partial{v}} \right\vert dv du $
$\iint\limits_{\sum} f\cdot d\sigma = \int_0^1 \int_0^{2-2u} (\frac67 u^2 + \frac{3uv}{7} + \frac67 -\frac67 u -\frac37 v )\frac72 dv du $
$\iint\limits_{\sum} f\cdot d\sigma = \frac74 $
But the answer provided is $\frac{15}{4}.$How is that? Where are we wrong in the computation of answer?
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