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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?

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I recommend that you project into the $xz$-plane, and use symmetry to consider half the cylinder ($y=\sqrt{2z-z^2}$). General advice: To find the region of integration, you want to project the intersection of the two surfaces; you do this by eliminating the remaining variable—in this case, $y$.

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  • $\begingroup$ Thank you very much! Your general advice cleared up my confusion. Now I can solve it: $S=2\int_0^2\int_{-\sqrt{2z}}^{\sqrt{2z}}\sqrt{[(1-z)^2/(2z-z^2)]+1}\,dx\,dz$ $\endgroup$ – Gary Aug 26 '13 at 12:55
  • $\begingroup$ Glad I could help, Gary. Feel free to accept the answer so that your question will not stay on the "unsolved" list. :) $\endgroup$ – Ted Shifrin Aug 26 '13 at 13:00
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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 $$\omega=\int_{-\pi}^\pi\ell(\phi)\ d\phi=8\int_{-\pi}^\pi\cos{\phi\over2}\ d\phi=16\ .$$

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  • $\begingroup$ Your answer is much simpler, but the question is an exercise of double integral, so I prefer to use that method for the purpose of learning. Anyway thanks! $\endgroup$ – Gary Aug 26 '13 at 13:02

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