Area of plane $2x+y+2z=16$ in the cylinder $x^2+y^2=16$ I'm having trouble making sure I'm getting the correct answer so I'll walk through it and hopefully someone can point out if it is correct.
$$2x+y+2z=16 \rightarrow z=8-x-{y\over2}$$ $$A(s)={\int \int}_D \sqrt{1+({\partial z\over \partial x})^2+({\partial z\over \partial y})^2}={\int \int}_D \sqrt{9\over4}dA={3\over2}{\int \int}_DdA$$ Then I transform into cylindrical: $${3\over2} { {\int_0^{2\pi}} {\int_0^4} rdrd\theta}=24\pi$$ I think the biggest problem I'm having justifying this is that I'm not sure I can go to cylindrical cordanites where I did, is that valid? Is the answer correct?
Thanks for any help!
Your $dS$ is a constant so your integral is simply the area of the circle time $dS$
$$A= \int \int _D dS = \int \int _D 3/2 dA = 3/2(16\pi ) = 24\pi $$