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?

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- Apr 17th 2008, 05:27 PMapw235@nyu.eduArea of the Surface
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? - Apr 17th 2008, 05:49 PMJen
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! :) - Apr 17th 2008, 06:05 PMapw235@nyu.edu
i think its $\displaystyle z=10-6cos\theta-15sin\theta$, but i'm confused as to how to use this, since we only have an equation for dS in cartesian coordinates => $\displaystyle \sqrt{(1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2)}$

- Apr 17th 2008, 06:15 PMJenI should have read the title to the post
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!! :)