Hello -
I am to find the part of the plane 2x+5y+z=10 that lies inside the cylinder $\displaystyle x^2+y^2=9$
I'm not sure how to start, and this is all i need help with. Do i parametrize the cylinder?
Using cylindrical coordinates, we know that
$\displaystyle x=rcos\theta$ and
$\displaystyle y=rsin\theta$
$\displaystyle 0\le r \le 3$
and $\displaystyle 0\le\theta\le\ 2 \pi$
Substituting all this for x and y after isolating z in the equation of the plane we get,
$\displaystyle z=10-2(3)cos\theta -5(3)sin\theta$
which simplifies to
$\displaystyle z=10-6cos\theta-15sin\theta$
Hope that helps!
Right, now that I have figured out what we are doing.
$\displaystyle \int \int 1 dS$ is the surface area.
so we get
$\displaystyle \int \int \sqrt{1+(-2)^2+(-5)^2}dA$
since we want only the area inside the circle we convert the integral to polar coordinates
$\displaystyle \sqrt{30} \int \int dA=\sqrt{30} \int_{0}^{2 \pi} \int_{0}^{3}rdrd\theta$
Hope that helps!!