1. Area of the Surface

Hello -

I am to find the part of the plane 2x+5y+z=10 that lies inside the cylinder $x^2+y^2=9$

I'm not sure how to start, and this is all i need help with. Do i parametrize the cylinder?

2. Using cylindrical coordinates, we know that
$x=rcos\theta$ and
$y=rsin\theta$

$0\le r \le 3$
and $0\le\theta\le\ 2 \pi$

Substituting all this for x and y after isolating z in the equation of the plane we get,

$z=10-2(3)cos\theta -5(3)sin\theta$

which simplifies to

$z=10-6cos\theta-15sin\theta$
Hope that helps!

3. i think its $z=10-6cos\theta-15sin\theta$, but i'm confused as to how to use this, since we only have an equation for dS in cartesian coordinates => $\sqrt{(1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2)}$

4. I should have read the title to the post

Right, now that I have figured out what we are doing.

$\int \int 1 dS$ is the surface area.

so we get

$\int \int \sqrt{1+(-2)^2+(-5)^2}dA$

since we want only the area inside the circle we convert the integral to polar coordinates

$\sqrt{30} \int \int dA=\sqrt{30} \int_{0}^{2 \pi} \int_{0}^{3}rdrd\theta$

Hope that helps!!