Find the surface area of the portion of the cone $z^2=x^2+y^2$ that is inside the cylinder $z^2=2y$.I know how to write parametric equation of simple surfaces and calculate area,but how do we find "area element" in situation like this.
I have solved this question but a few more question have come in my mind.
- Can the area of surface formed by intersection be calculated through surface integrals?(means is there any general method to write their parametric equation)
- Is surface area of inside equal to area outside?(means area inside a ball and outside it)
- If yes then can 1. be solved by calculating area of two surfaces separately and then adding them?
Best Answer
In a situation like this, one usually uses symmetry to focus on the upper half of the cone, with equation $z=\sqrt{x^2+y^2}$. The area element in this case is simply $$\sqrt{1+|\nabla z|^2}\,dx\,dy = \sqrt{1+\frac{x^2+y^2}{x^2+y^2}}\,dx\,dy = \sqrt{2}\,dx\,dy \tag1$$ Integration happens over the region $x^2+y^2\le 2y$, which is a closed disk of radius $2y$.
Just kidding, it's a disk of radius $1$ with center at $(1,0)$. So the integral is $\sqrt{2}$ times $\pi 1^2$. And yes, this was a half of original surface, so the final answer is $2\sqrt{2}\pi$.
That's the only [calculus] way to calculate surface area: $$A(S)=\iint_A 1\,dS = \iint |r_u\times r_v|\,du\,dv$$ The above is a special case, with parameters $u$ and $v$ being $x,y$.
I've no idea what inside and outside mean here. But the area above $z=0$ is equal to the area below it, by symmetry.
That's what I did above.