So I stumbled across an exam question where it gives a surface area where:
$$ S = \{(x, y, z) : x, y, z ≥ 0, 2x + y + 2z = 4\}. $$
It then asks to sketch this surface area and we can see it's a line then it goes on to say:
Find the equation of the line segment where $S$ intersects the $xy$-plane.
So what I'm thinking of doing is that when the line intersects the $xy$-plane doesn't $z = 0$?
and also to:
Evaluate $\iint_{S} x + y + z\,dS $
Hint: Parametrize $S$ in the form $(x, y, f(x, y)), (x, y) \in \mathbb{R}$.
I'm quite new to all this and am finding it hard to understand what to do. If anyone could help with these two parts I would really appreciate it!
Best Answer
At $z=0$, $2x+y=4 \implies y=4-2x$ this will help for the integration.
Note that the $x$, $y$ and $z$ intercepts are $2$, $4$ and $2$ respectively.
\begin{align*} \boldsymbol{r} &= \left( x,y,2-x-\frac{y}{2} \right) \\ \boldsymbol{r}_{x} &= (1,0,-1) \\ \boldsymbol{r}_{y} &= \left( 0,1,-\frac{1}{2} \right) \\ \boldsymbol{r}_{x} \times \boldsymbol{r}_{y} &= \left( 1,\frac{1}{2}, 1 \right) \\ dA &= |\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}| \, dx \, dy \\ &= \frac{3}{2} \, dx \, dy \\ A &= \int_{0}^{2} \int_{0}^{4-2x} \frac{3}{2} \, dy \, dx \\ &= \int_{0}^{2} \frac{3}{2} (4-2x) dx \\ &= \left[ 6x-\frac{3x^{2}}{2} \right]_{0}^{2} \\ &= 6 \end{align*}