# Find the Area of the Surface That Lies inside the Cylinder

Find the area of the surface. The part of the plane $5x+2y+z=10$ that lies inside the cylinder $x^2 + y^2 = 25$.

I know the double integral formula for finding surface area and how to find the partial derivatives that go into the formula. But how do I find the limits of integration? I'm stuck on finding the intersection of these two curves.

The surface can be parametrized as follows: $$x=x,\quad y=y, \quad z=10-5x-2y\quad (x,y)\in D$$ with $$D=\{(x,y)\;|\; x^2+y^2\le 25 \}$$ It follows that the wanted area equals $$A=\iint_D ||r_x\times r_y || \;dA =\iint_D \sqrt{5^2+2^2+1^1} \;dA =\sqrt{30}\;A(D)=\sqrt{30}\;\pi5^2$$

HINT : can you write $x$ and $y$ in terms of $\cos(t)$ and $\sin(t)$ ? then find $z(t)$ ? then find the area from the line integral ?

You can use polar coordinates:

$x=r\cos t;\ y=r\sin t$ so that $\vec r=r\cos t\vec i+r\sin t\vec j+(10-5r\cos t-2r\sin t)\vec k$.

Then,

$\vec r_r=\cos t\vec i+\sin t\vec j-(5\cos t+2\sin t)\vec k$

$\vec r_t=-r\sin t\vec i+r\cos t\vec j+(5r\sin t-2r\cos t)$

$\vec r_r\times \vec r_t =5r\vec i-2r\vec j+r\vec k\Rightarrow \vert \vec r_r\times \vec r_t\vert=r\sqrt{30}$.

Putting this all together, we can do the surface integral:

$\int_Sd\vec S=\int \sqrt{30}rdA=\sqrt{30}\int_{0}^{2\pi }\int^{5}_0 rdrdt=25\pi\sqrt{30}.$