How should I setup the integral to calculate surface area

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?

• 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.) – saulspatz Aug 23 '19 at 15:32

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.

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

• Actually,the cylinder is at centre $(0,1)$, so I think the parametric coordinates ad the corresponding integral will change accordingly – sat091 Aug 24 '19 at 18:02
• It was just a typo in this line. The rest is correct. The displaced cylinder is in polar coordinates $r = 2 \cos(\theta)$. – Franco M. Aug 24 '19 at 18:26
• Which part is not clear for you? – Franco M. Aug 24 '19 at 18:33
• 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. – sat091 Aug 24 '19 at 19:23
• 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. . – sat091 Aug 24 '19 at 20:09