Calculate surface area of $f(x+y+z)$

Question

Calculate the following surface integral: $$\iint_{\sigma}(x+y+z)\mathrm{d}S$$ When $\sigma$ is the plane $x+y=1$ (x,y positive) between $z=0$ and $z=1$.

For some reason I don't manage because I can't express $z$ in terms of $x$ and $y$.

Answer 1

There are two simplifications to be made. Since we are on the surface $x+y=1$ we have that

$$\iint\limits_\sigma x+y+z\:dS = \iint\limits_\sigma 1 + z\:dS$$

Second, the surface is a rectangle with side lengths $1$ and $\sqrt{2}$. Since the function only varies in the $z$ direction, we can say that

$$\iint\limits_\sigma 1+z\:dS = \sqrt{2}\int_0^11+z\:dz = \frac{3}{\sqrt{2}}$$

Answer 2

While this is a more straightforward case but given your question that you are not able to parametrize as $z =f(x, y)$, you always do not have to express $z$ in terms of $x, y$. We can also express as $x = f(y,z)$ for example. It depends on the surface and in which plane, the projection would make more sense.

For this question, let's take the projection in $YZ$ plane.

Here the given surface is $x+y=1, x, y \geq 0$, so we can express it as $x = f(y) = 1-y$.

So $\sqrt{1+ (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2} \ = \sqrt2$

Also integrand $x+y+z = 1-y+y+z = 1+z$

So the surface integral is

$\displaystyle \int_0^1 \int_0^1 \sqrt2 \ (1+z) \ dy \ dz = \frac{3}{\sqrt2}$