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?
First let's resume what we have:
$y^2 + z^2 = 2 z$ 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.
$x^2 = y^2 + z^2$ 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:
$$y = r \cos(\theta)$$
$$z = r \sin(\theta)$$
$$x = x$$
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.
$$ S = 4 \int^2_{x = 0} \int^{\pi/2}_{\theta = \arcsin(x/2)} 2 \, d\theta \, dx = 8 \int^2_{x = 0} \! \frac{\pi}{2} - \arcsin(x/2) \, dx = {\color{red}{16}}$$