How do you find the surface area of the part of the plane $\displaystyle 3x+y+z=4$ that lies inside the cylinder $\displaystyle x^2+y^2=4$
Let $\displaystyle f(x,y) = z = 4 - 3x - y$
Then, $\displaystyle f_x = -3$ and $\displaystyle f_y = -1$. The cylinder intersects the $\displaystyle xy$-plane at the circle of radius 2 centered at the origin, that is, over the region where $\displaystyle - \sqrt{4 - x^2} \le y \le \sqrt{4 - x^2}$ and $\displaystyle -2 \le x \le 2$.
The surface area is given by $\displaystyle A = \int_{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt {4 - x^2}} \sqrt{(f_x)^2 + (f_y)^2 + 1}~dy~dx = \int_{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt {4 - x^2}} \sqrt{11}~dy~dx $ $\displaystyle = \int_0^{2 \pi} \int_0^2 \sqrt{11}r~dr ~d \theta$
I leave the rest to you