I was doing this question,
Evaluate the surface integral $\int \int_S 3z(d\sigma)$, where the surface $S$ is the portion of the plane $2x+y+z=2$ lying in the first octant.
The way I approached this question was that I took small elemental triangles being projected from the XY-plane, like this,
Here I multiplied the length of hypotenuse $\sqrt {x^2+y^2}$ with $dz$ and then substituted $x=\frac{2-z}{2}$ and $y=2-z$, and the resulting expression is,
$S=3\int_0^2z\sqrt{\frac {5(2-z)^2}{4}}(dz)$
However, after evaluating the integral, I am getting an incorrect answer of $2\sqrt 5$. However, the correct answer is $2\sqrt 6$.
So please help me spot my mistake. Any help would be appreciated!
Best Answer
The integral is indeed $2\sqrt6$, derived here using this well-known formula. With $z=f(x,y)=2-2x-y$, we have
$$\begin{align} \iint_S 3z\,\mathrm d\sigma&=\int_0^1\int_0^{2-2x} 3(2-2x-y)\sqrt{1+{f_x}^2+{f_y}^2}\,\mathrm dy\,\mathrm dx\\[1ex] &=\int_0^1\int_0^{2-2x}3(2-2x-y)\sqrt{1+(-2)^2+(-1)^2}\,\mathrm dy\,\mathrm dx\\[1ex] &=3\sqrt6\int_0^1\int_0^{2-2x}(2-2x-y)\,\mathrm dy\,\mathrm dx\\[1ex] &=3\sqrt6\times\frac23=\boxed{2\sqrt6} \end{align}$$