Example:
Find the area of the surface of the cap cut from the paraboloid $z=12-x^2-y^2$ by the cone $z=x^2+y^2$.
I've seen some approaches of taking the magnitude of the cross product of the two partial derivatives of the equation of the parabaloid, but I still don't know how the equation of the cone plays into this problem. What would be the best way to approach this?
Thanks!
It turns out both of these surfaces are paraboloids. We'll refer to the one with equation $z=x^2+y^2$ as bottom for now. What we want to do first is find the intersection of these two surfaces, which is the only place where we will use the bottom. Doing so gives the equation for the integrating region, D.
$12-x^2-y^2=x^2+y^2 $
Since we're done with the bottom, 'paraboloid' now refers to the top one.
Next, we can make use of a vector that can trace any surface
r$(x, y, z) = xi + yj + zk$
We specifically want it to trace the top portion of the paraboloid by making it a function only of 2 variables, x and y
Let $f(x,y)$ be the surface to trace, in this case, the paraboloid
r$(x,y)=xi + yj + f(x,y)k$
Now, taking the partials with respect to x and y form a tangent plane to the surface at each point on that surface. Their cross product comes into play since the magnitude of their cross product is equal to the area of the parallelogram they form. This is what we are interested in integrating over.
$\frac{\partial r}{\partial x}$ = $i + \frac{\partial f}{\partial x}k$
$\frac{\partial r}{\partial y}$ = $j + \frac{\partial f}{\partial y}k$
n $=\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y}$
When evaluating vector valued surface integrals $($given F$(x,y,z)$ as opposed to $f(x,y))$, the orientation of n matters, but we are only concerned with its magnitude, so the ordering does not matter.
n $=-\frac{\partial f}{\partial x}i-\frac{\partial f}{\partial y}j+k$
So
|n|$=\sqrt{1+\frac{\partial f}{\partial x}^2+\frac{\partial f}{\partial y}^2}$
Then from here, the surface area of the cap is just an integral away
$\iint_S\mathrm{dS} = \iint_D$|n|$\mathrm{dA}$
Since D will turn out to be a circular region in this case, it is always worth considering performing the integration in polar coordinates