I've looked over numerous surface integral questions, but I just cannot understand how the projection works.
I'm given $A: (x+y^2)i – (2x)j + (2yz)k$.
$S$ is the surface of the plane $2x+y+2z=6$ in the first octant.
How do we know on which plane the projection will work best and how the limits of integration work. The method I've been taught is:
$\iint A \cdot \hat{n} \ ds$, where $\hat{n}$ is the unit normal vector perpendicular to the surface, and $ds = \frac{dx\,dy}{\hat{n} \cdot \hat{k}}$, if the projection is onto the $xy$ plane.
Can someone please solve this question with this method and clear my confusion on limits and where and how to take the projection.
Thanks a lot!
Best Answer
You start by finding the unit normal to your surface $F(x,y,z)=2x+y+2z-6=0$ (which is clearly constant in this case): $$ \mathbf{n}=\frac{\nabla F}{\|\nabla F\|}=(2/3,1/3,2/3). $$ Given that it is a graph, we can use $x$ and $y$ as parameters. Next, you find the dot product $\mathbf{A}\cdot\mathbf{n}$, in terms of $x,y$: $$ \mathbf{A}\cdot\mathbf{n}=2(x+y^2)/3-2x/3+4y(-x-y/2+3)/3=4y-\frac{4xy}{3}. $$ Finally, your area element is $dS=\frac{3}{2}dxdy$ (see the surface area computation). You then compute the double integral $$ \frac{3}{2}\iint_\Delta(4y-\frac{4xy}{3})\,dxdy $$ where $\Delta$ is as before, the triangle cut out by the line $2x+y-6=0$ in the first quadrant: $$ \frac{3}{2}\iint_\Delta(4y-\frac{4xy}{3})\,dxdy=\frac{3}{2}\int\limits_0^3\left(\int\limits_0^{-2x+6}\left(4y-\frac{4xy}{3}\right)\,dy\right)\,dx=\frac{3}{2}\int\limits_0^3\left(\left(4-\frac{4x}{3}\right)\int\limits_0^{-2x+6}y\,dy\right)\,dx $$ $$ =\frac{3}{2}\int\limits_0^3\left(-\frac{8}{3}x^3+24x^2-72x+72\right)\,dx $$ which is a simple polynomial integral. The final result is $81$.