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.
Best 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: