Here's the question:
Find the surface area of the part of the plane $2x+y+2z=16$ bounded by the surfaces $x=0$, $y=0$ and $x^2+y^2=64$.
So, I know I have to parameterize the surface $S:\mathbf{x}=\mathbf{x}(u,v)$ and I currently have it parameterized as $\mathbf{x}(u,v)=(\sqrt{32}\cos{u},\sqrt{32}\sin{u},v)$ because we're dealing with a circle of radius $8$ in the first quadrant (octants 1 and 5). If my parameterization is correct, then $$\left\lVert\frac{\partial\mathbf{x}}{\partial u}\times \frac{\partial\mathbf{x}}{\partial u}\right\rVert=64.$$
I'm still confused about the bounds though. Obviously, $0\le u\le\pi/2$ but what about $v$?
Help!
Thank you :)
The parametrization of $2x+y+2z=16$ bounded by the surfaces $x=0$, $y=0$ and $x^2+y^2=64$ can be naively parametrized by: $$ S: \mathbf{\Phi}(x,y) = \left(x,y , \frac{16-2x-y}{2}\right), $$ now we want to find the bounds on $x$ and $y$: for a point on $x^2+y^2=64$ with a fixed $x$-coordinate, $0\leq x\leq 8$: $y = \sqrt{64-x^2}$. Therefore for a point on this surface: $$ 0\leq x\leq 8, \\ 0\leq y \leq \sqrt{64-x^2}. $$
Now the integral to compute the area is: $$ A = \iint_S 1 \,dS = \int^8_0\int^{\sqrt{64-x^2}}_0 \left|\frac{\partial \mathbf{\Phi}}{\partial x}\times \frac{\partial \mathbf{\Phi}}{\partial y}\right| dydx = \int^8_0\int^{\sqrt{64-x^2}}_0\frac{3}{2}dydx = \frac{3}{2}\int^8_0 \sqrt{64-x^2}\,dx. $$ This is like computing a slanted flat face's area from $\{x\geq 0,y\geq 0: x^2+y^2\leq 64\}$ (it is highly recommended you sketch the graph yourself).