# Surface area of intersection of surfaces

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.

1. Can the area of surface formed by intersection be calculated through surface integrals?(means is there any general method to write their parametric equation)
2. Is surface area of inside equal to area outside?(means area inside a ball and outside it)
3. If yes then can 1. be solved by calculating area of two surfaces separately and then adding them?
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The area element is still that of the cone. It's just that your limits of integration will be defined by the intersection curve. –  Rahul Feb 8 '13 at 15:03
Thanks :),i was mistaking it for area of surface formed by intersection.Can we do that? –  shivam Feb 8 '13 at 18:43
What do you mean by "surface formed by intersection"? The intersection is not a surface, it is a curve, marked in red in your diagram. –  Rahul Feb 9 '13 at 8:24
For example an hemisphere $x^2+y^2+z^2=4$ and paraboloid $z-1=x^2+y^2$ –  shivam Feb 9 '13 at 13:20
I give up. ${}$ –  Rahul Feb 9 '13 at 16:07

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 $2\sqrt{2}\pi$.

Can the area of surface formed by intersection be calculated through surface integrals?(means is there any general method to write their parametric equation)

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 $x,y$.

Is surface area of inside equal to area outside?(means area inside a ball and outside it)

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.

If yes then can 1. be solved by calculating area of two surfaces separately and then adding them?

That's what I did above.

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