The question is:
Find the surface area of the part of the cylinder $$y^2+z^2=2z$$ that is cut off by the cone $$x^2=y^2+z^2$$
(answer=$16$)
This question is from a section that discuss the calculation of the area of the surface $z=f(x,y)$ that lies over the region $R$, by using the double integral: $$S=\iint_R\sqrt{f_x^2+f_y^2+1}\,dx\,dy$$
So we are expected to use double integral to get the answer.
Notice that the question is asking for the area of the part of the cylinder, not the intersection area between the cylinder and the cone.
My problem is: how to define $R$?
- if $R$ lies on the $yz$-plane, then the surface of the cylinder does not lie over $R$, so we can't use the above formula.
- if $R$ lies on the $xy$-plane, then the equation of cylinder gives $z=1\pm \sqrt{(1-y^2)}$, which is not a function.
- if $R$ lies on the $xz$-plane, then how can we get the equation of $R$? seems that it is not correct to just set $y=0$ in $x^2=y^2+z^2$ to get $x=\pm z$, because it removes the information that the cylinder is cut off by a cone.
So, do you have any idea on how to solve it?
Best Answer
The simplest way to compute the requested area would be the following:
The axis of the cylinder $S$ is parallel to the $x$-axis, and $S$ intersects the $(y,z)$-plane in the circle $y^2+(z-1)^2=1$ of unit radius. A parametric representation of this circle is given by $$y=\sin\phi,\quad z=1+\cos\phi\qquad(-\pi\leq \phi\leq\pi)\ .$$ Through each point on this circle passes a stalk $s_\phi$ parallel to the $x$-axis whose ends are determined by $$x^2=y^2+z^2=2(1+\cos\phi)\ .$$ It follows that $s_\phi$ has length $$\ell(\phi)=2\sqrt{2(1+\cos\phi)}=4\cos{\phi\over2}\qquad(-\pi\leq \phi\leq\pi)\ .$$ The area $\omega$ of the surface in question is therefore given by $$\omega=\int_{-\pi}^\pi\ell(\phi)\ d\phi=8\int_{-\pi}^\pi\cos{\phi\over2}\ d\phi=16\ .$$