Let D be a closed and bounded region on the xy-plane that does not contain the origin. Show that the surface area of the part on the cone $z =
\sqrt{x^2+y^2}$ lying above D is $\sqrt{2}a$, where $a$ is the area of D.
To find the surface area, I have to solve the double integrals:
$\int\int_DdS$. By partial differentiation, $f_x = \frac{x}{\sqrt{x^2+y^2}}$ and $f_y = \frac{y}{\sqrt{x^2+y^2}}$.
Hence, $\int\int_DdS = \int\int_D \sqrt{1+(f_x)^2+(f_y)^2}dA$ where $dA = |dxdy|$. However, I am not sure of how to set the upper and lower bound of the double integral given D. May I get some help with how to proceed?
Best Answer
You've already gotten to $\iint_D\sqrt{1+(f_x)^2+(f_y)^2}dA$, so you're on the home stretch. Insert and simplify $(f_x)^2+(f_y)^2$, remember that $\iint_D1\,dA=a$, and you're done.