# Surface integral - The area of a plane inside a cylinder

I'm having trouble with this question: Find the Surface Area of the part of the plane $x+2y+z=4$ that is inside the cylinder $x^2+y^2=4$.

I tried writing the surface like this:

$$(r\cos(\theta), r\sin(\theta), -r\cos(\theta)-2r\sin(\theta))$$ with $\theta \in [0,2\pi]$ and $r \in [0, 2]$.

I'm not sure about the $r$, should I have used the radius of the cylinder ($2$) instead?

I took the partial derivatives with respect to $r$ and $\theta$ to find the $R_u$ and $R_v$ seen in the formula:

$$\iint_S(1\cdot {\rm d}S) = \iint_D ||{\rm R_u\times R_v}||{\rm d}A$$

But I got $||{\rm R_u\times R_v}|| = r\sqrt{\sin^4\theta+5\cos^4\theta}$ and I believe that can't be right. Can anyone help me out here, how should I do it?

I struggled to do the MathJax stuff, sorry for the bad notation.

• This is a correct approach (although technically, the $z$ component should be $4-r\cos \theta-2r\sin\theta$). However, you didn't calculate the cross product correctly. The normal of the plane is constant, there should be no angle dependence, just $r\sqrt6$. Try it again with care. There is a lot of simplification possible when you do the cross product, because $\sin^2+\cos^2=1$. To avoid calculus, see my answer below. Commented Jun 2, 2014 at 9:32
• I really and honestly appreciate all the effort you put into answering me, I ended up solving it with the "more calculus" way, achieved the same result as you did! But anyways I've read your answer quite a few times and I believe that now I finally understand the reasons of how to do it and, specially, why it works. Thank you very much! Commented Jun 2, 2014 at 10:13

## 1 Answer

The intersection of a cylinder with a plane is an ellipse. Find the semiaxes of the ellipse and you get $$S=\pi ab$$

The minor semiaxis is always the same as the radius of the cylinder, in this case $b=r=2$.

The major semiaxis can be calculated from the angle between the plane and the cylinder axis. The angle the plane makes with the $z$ axis can be extracted from the plane normal, as the dot product gives us a cosine between two vectors, $\cos\alpha=(n_x,n_y,n_z)\cdot(0,0,1)=n_z$.

The normal of your plane can be read directly from the coefficient of the equation. An equation for a plane can be written as a dot product $\vec{n}\cdot\vec{r}=\rm const$, in your case $(1,2,1)\cdot(x,y,z)=4$. Rescale the normal to unit size and you get: $$\vec{n}=\frac{(1,2,1)}{\sqrt 6}$$ and as we demonstrated above, also by using the dot product, the $z$ component of the normal equals the cosine of the angle with the $z$ axis: $$\cos\alpha=\frac{1}{\sqrt 6}$$

If you draw the vertical cross section (the figure on the right), you can see a right triangle that relates the radius of the cylinder with the hypotenuse (the semiaxis):

$$a=\frac{r}{\cos\alpha}=2\sqrt 6$$ leading to the solution $$S=4\pi \sqrt 6$$

EDIT:

• Thank you for your kind answer but I don't follow you, sorry. I understand the idea and am able to visualize the ellipse but I can't understand the process you used to find $b$, I drew it but still, no success. I'm always really lost when it comes to the need of my knowledge in geometry. Commented Jun 2, 2014 at 9:03
• I added a sketch. Better? $b$ is the easiest, no matter how you cut a cylinder, at the narrowest, the cut is always as wide as the cylinder itself. The major axis depends only on the tilt of the cut (α in our case). Left sketch is sort of hand-drawn 3D, right is the vertical cross section. Commented Jun 2, 2014 at 9:17
• Sorry, I was thinking about $a$ not $b$, my bad. The images helped, but I still don't understand, I can't link the normal of the plane to $\cos\alpha$. To be true, I don't even understand how you got to the normal line. Commented Jun 2, 2014 at 9:31
• @LuanCristianThums I rewrote that part now. There are "textbook rules" how to read the normal from the coefficients in the equation, but I find that the vector notation solves all the problems at once. The cosine is then just the dot product of two unit vectors. Remember this for later, it is extremely useful. Commented Jun 2, 2014 at 9:38