Integrals with $e^z$

Question

I need help starting this question.

Calculate the integral of $$f(x,y,z)=e^z$$ over the portion of the plane $$x+2y+z=4$$ where $x,y,z$ is greater than or equal to 0.

Answer 1

The surface integral in which the surface S is given by $z=g(x,y)$ is $\iint_Sf(x,y,z)dS=\iint_Df(x,y,g(x,y))\sqrt{(\frac{\partial{g}}{\partial{x}})^2+(\frac{\partial{g}}{\partial{y}})^2+1}dA$
In this case, $x=4-2y-z$
We can set the integral to $$\iint_Se^zdS=\iint_De^z\sqrt{(\frac{\partial{g}}{\partial{y}})^2+(\frac{\partial{g}}{\partial{z}})^2+1}dA=\iint_De^z\sqrt{(-2)^2+(-1)^2+1}dA=\sqrt6\iint_De^z$$
We can also think of the problem in this way:
We have some freedom in choosing which coordinate plane to project the surface S onto. Let’s project S into the yz-plane. Then the projection vector is p = i.
$g(x,y,z)=x+2y+z$
$\nabla{g}=<1,2,1>, |\nabla{g}|=\sqrt6,|\nabla{g}\cdot p|=|\nabla{g}\cdot i|=\sqrt6$

Now we have $\sqrt6\iint_De^z$ and we need to find the limits. Set x=0, we can get z=4-2y

Set x=0, z=0, we can get thee upper bound for y, which is 2

Thus the final integral is $$\sqrt6\int_0^2\int_0^{4-2y}e^zdzdy$$