Surface integral of function over a triangle

Question

I'm working through some problems in a textbook and I wasn't sure how to do this

I have a function $f(x,y,z) = xyz$,and a region $S$ which is the triangle with vertices (3,0,0), (0,2,0) and (0,0,6) and I want to find the surface integral

$$\iint_S f(x,y,z) dS$$

I understand the theory behind this - I take two vectors in the region $S$, say (-3,2,0) and (-3,0,6) and find the cross product, which gives me a normal vector and then I parametrize $f(x,y,z)$ and integrate over $S$ which has bounds expressed in terms of my new coordinates.

I'm having trouble understanding how to paramatrize the region and the function - how do I do that?

Thanks.

Answer 1

I'm not sure why you want to take the cross product. Given your two vectors $u=(-3,2,0)$ and $v=(-3,0,6)$, and a point $A=(3,0,0)$, the points in the plane that contain your triangle are the points $P(\alpha,\beta) = A + \alpha u + \beta v$, for $\alpha,\beta\in\mathbf R$. Now you just have to find the bounds on $\alpha$ and $\beta$ for which the point $P(\alpha,\beta)$ stays in the triangle $S$ (which should be simple given your judicious choice of vectors), and you have yourself a parametrization of $S$.

I'm also not sure what parametrizing the function $f$ means. You just have to rewrite $f(x,y,z)$ in terms of $\alpha$ and $\beta$, and integrate.

Answer 2

The triangle $S$ lies in the plane $\pi$ with equation $${x\over3}+{y\over2}+{z\over 6}=1\ ,$$ or $z=6-2x-3y$. Let $$S':=\left\{(x,y)\ \bigg|\ 0\leq x\leq 3,\ 0\leq y\leq 2-{2x\over3} \right\}$$ be the projection of $S$ onto the $(x,y)$-plane.

The normal vector of $S$ is parallel to $\left({1\over3},{1\over2},{1\over 6}\right)$. Denoting the angle between this vector and the $z$-axis by $\alpha$ we have $$\cos\alpha=(0,0,1)\cdot{1\over\sqrt{14}}(2,3,1)={1\over\sqrt{14}}\ .$$ This angle $\alpha$ is also the angle of inclination of the plane $\pi$ versus the $(x,y)$-plane. Therefore the scalar surface element ${\rm d}\omega$ on $S$ and the area element ${\rm d}(x,y)$ on $S'$ are related by $${\rm d}\omega={1\over\cos\alpha}{\rm d}(x,y)=\sqrt{14}\ {\rm d}(x,y)\ .$$ It follows that $$\eqalign{\int_S f\ {\rm d}\omega&=\sqrt{14}\int_{S'} f(x,y,6-2x-3y)\ {\rm d}(x,y)\cr &= \sqrt{14}\int_0^3\ \int_0^{2-2x/3} f(x,y,6-2x-3y)\ dy\ dx\ .\cr}$$