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Find the surface area of the part of the surface $y^2 +z^2 =2z$ cutoff by the cone $x^2 = y^2+z^2$

So the given surface is actually a circle with radius $1$ and centre $(0,1)$ in y-z plane,but I am little confused on how to setup the integral to calculate the surface area .

Can anyone tell me how to should I setup the integral for this question?

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  • $\begingroup$ Are the equations typed correctly? As it is, I think the intersection is the surface $x^2=2z$, which is unbounded. (A parabolic cylinder, I think you'd call it.) $\endgroup$ – saulspatz Aug 23 '19 at 15:32
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First let's resume what we have:

  1. $y^2 + z^2 = 2 z$ refers to a cylinder with the middle axis in $x$-direction and the center at $y=0$ and $z=1$ and a radius of 1.

  2. $x^2 = y^2 + z^2$ is a double cone centered at the origin with the axis along $x$-direction

It is useful to transform into polar coordinates in this case:

$$y = r \cos(\theta)$$

$$z = r \sin(\theta)$$

$$x = x$$

  1. $y^2 + z^2 = 2z \Leftrightarrow r = 2 \sin(\theta)$
  2. $x^2 = y^2 + z^2 \Leftrightarrow x^2 = r^2$

Due to symmetry ($y$-$z$-plane and $x$-$z$-plane are mirror planes) we can integrate in the positive octant and multiply the surface are times 4.

For the surface integral we will need to integrate over $x$ and $\theta$. You can imagine that one evaulates the arc length of a circle and the integrate along $x$-axis to get a cylinder. What we need for the boundaries is a function $\theta = f(x)$. The boundaries for $x$ will be numeric values.

The boarders for $x$ are easily obtained. Just imagine a side view:

$$0 \leq x \leq 2$$

The connection between $x$ and $\theta$ requires some trigonometry. If you look from above along the $x$-axis, you will see two circle: one that refers to the cone at has a radius of $r=x$ and a second displace circle with $r=1$. From this follows:

$$\arcsin \Big(\frac{x}{2} \Big) \leq \theta \leq \frac{\pi}{2}$$

The surface area for polar coordinates can be calculate with:

$$S = \int \int \! \sqrt{r^2 + \Big( \frac{\partial r}{\partial \theta} \Big)^2} \, d\theta \, dx$$

Finally, we will have to solve.

$$ S = 4 \int^2_{x = 0} \int^{\pi/2}_{\theta = \arcsin(x/2)} 2 \, d\theta \, dx = 8 \int^2_{x = 0} \! \frac{\pi}{2} - \arcsin(x/2) \, dx = {\color{red}{16}}$$

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  • $\begingroup$ Actually,the cylinder is at centre $(0,1)$, so I think the parametric coordinates ad the corresponding integral will change accordingly $\endgroup$ – sat091 Aug 24 '19 at 18:02
  • $\begingroup$ It was just a typo in this line. The rest is correct. The displaced cylinder is in polar coordinates $r = 2 \cos(\theta)$. $\endgroup$ – Franco M. Aug 24 '19 at 18:26
  • $\begingroup$ Which part is not clear for you? $\endgroup$ – Franco M. Aug 24 '19 at 18:33
  • $\begingroup$ I have a little doubt for bounds of $\theta$ ,I think they should vary from $\arccos(x/2)$ because we are setting up the bound between $x=r$ and $x=2cos\theta$,also can you please explain how you calculate the bounds on $x$,I am having some trouble in it. Thank you. $\endgroup$ – sat091 Aug 24 '19 at 19:23
  • $\begingroup$ I searched this site and discovered the exact same question was asked here :math.stackexchange.com/questions/475925/… 6 years ago, as you can see the answer given there is 16. I think we both are making some mistake in setting up the bounds of this integral. . $\endgroup$ – sat091 Aug 24 '19 at 20:09

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