# Find the surface area of portion of the plane that is inside the cylinder.

I am given the plane $x + y + z = 1$ and the cylinder $x^2 + y^2 = 4,$ and have to find the surface area of portion of the plane that is inside the cylinder.

I am very confused with this. I tried writting the intersection of the two surfaces as a parametric curve and got: $$\mathbf{r} (t) = (2 \cos{t}, 2 \sin{t}, 1 - 2 \cos{t} - 2 \sin{t}).$$ The plan was then to calculate the surface area enclosed by this curve, but I don't know how to do this. I know there are formulas for doing that when the curve has parametric equation of the form $\mathbf{s}(t) = (x(t), y(t)),$ but not in my case. How can I do it?

• You're looking for Stoke's theorem (or as it will be very likely be called in this particularly case, Green's theorem). Commented May 19, 2014 at 17:57

## 2 Answers

$z=1-x-y$ so set $f(x,y)=1-x-y$ $$S=\int\int_D\sqrt{f_x^2+f_y^2+1}dxdy$$

$$S=\int\int_D\sqrt{3}dxdy$$

$$S=\sqrt{3}\int\int_Ddxdy$$

$$S=\sqrt{3}(4\pi)$$

$$S=4\sqrt{3}\pi$$

Note: $D$ is the region enclosed by the circle $x^2+y^2=4$ in $x,y$ plane.

Your equation for the curve is correct.

$$\mathbf{r} (t) = (2 \cos{t}, 2 \sin{t}, 1 - 2 \cos{t} - 2 \sin{t}).$$

You have only to compute the extrema of the z component, one lies at $x=y$ where $t = \frac{\pi}{4}$ to find the semi-major axis $a = 2\sqrt{3}$ and the value at $x=-y$ at $t = \frac{\pi}{4} + \frac{\pi}{2}$ to find the the semi-minor axis $b = 2$ , which is enough information to calculate the area of the ellipse: ${a} {b} \pi = 4\sqrt{3} \pi$ .

Incidentally the center of the ellipse lies at (0,0,1) , you will need this point to compute a and b .