# Evaluate the surface integral $\iint\limits_{\sum}f\cdot d\sigma$

• MHB
• WMDhamnekar
In summary, to evaluate the surface integral on the given plane, we need to parametrize $\sum$ using the given restrictions and outward unit normal. However, the computation of the answer provided is incorrect, and the correct answer is $\frac{7}{4}$, not $\frac{15}{4}$.

#### WMDhamnekar

MHB
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?

Last edited:
Remember that $$\mathbf{n} \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right\|\,du\,dv = \left(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right)\,du\,dv.$$

With your parametrization, $$\mathbf{n}\,d\sigma = \begin{pmatrix} 3 \\ \frac32 \\ 1 \end{pmatrix}\,du\,dv$$ and hence $$\int_\Sigma \mathbf{f} \cdot \mathbf{n} \,d\sigma = \int_0^1 \int_0^{2(1-x)} 3x^2 + \frac32 (x-1)y + 3(1 - x) \,dy\,dx$$ for which I also get $\frac 74$.