Find the surface area of the part of the plane $2x+y+2z=16$ bounded by the surfaces $x=0, y=0$ and $x^2+y^2=64$
I recognised this as the region in the first quadrant in the $xy$ plane of $x^2+y^2=64$. After parametrising where i let $v=x$ and $u=y$, my double integral was $\iint_V \,du\,dv$ where $u$ ranged from $-\sqrt{64-x^2}\cdots\sqrt{64-x^2}$ and $v$ ranged from $0\cdots8$. Evaluating this gave $64\pi$, which is incorrect. Any suggestions?
Notice the constraints you are given create a domain on the $xy$ axis. Given this, consider the height of plane: $$z = \frac{16-2x-y}{2}.$$ We can now use the following result (see here) to calculate the surface area of some surface $f(x,y)$ over a domain $D$ in the $xy$ plane:
The surface area of $f(x,y)$ over a domain $D$ is given by $$S = \int \int_D \sqrt{[f_x]^2+[f_y]^2+1} dA$$ where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively.
Given how in our problem $f(x,y)= \frac{16-2x-y}{2}$, the partial derivatives are $f_x = -1$ and $f_y= -1/2$. Notice the integrand is a constant (i.e. $\sqrt{1+1/4+1}=3/2$). Now, draw a picture of what $D$ looks like in the $xy$ plane. It is a quarter of a circle with radius 8. Given this information, $\int \int_D dA$ is quarter of the area of a circle with radius 8. Therefore,
$$S = \int \int_D \sqrt{[f_x]^2+[f_y]^2+1} dA = \frac{3}{2}\int \int_D dA = \frac{3}{2}\cdot \frac{\pi 8^2}{4} = 24\pi $$