# Surface Integral - Intersection of cone and circle

I am now looking at the following exercise: Calculate: $\int \int_S (2z^2 - x^2 - y^2) dS$ where $S$ is $z=\sqrt{x^2+y^2}$ intersected with $x^2 + y^2 =2x$ (i.e. $(x-1)^2 + y^2 =1$ ) .

Now, as far as definitions: $\int \int_S (2z^2 - x^2 - y^2) dS = \int \int_D f(x(u,v),y(u,v),z(u,v)) ||\phi_u \times \phi_v || dudv$ where $\phi:D \to \mathbb{R}^3$ is the parameterization of $S$ .

But, how can I parameterize $S$ ? It is the intersection of the cone and the cylinder, but how can I do it?

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the circle is x-y plane, how does it intersect with cone, also i think the intersection of two surfaces will give a curve unless they are tangential on some region. –  Santosh Linkha Jan 16 '13 at 20:39
We are probably dealing with the surface that its $z$ coordinate is $z= \sqrt{x^2+y^2}$ and $x^2+y^2=2x$ ... –  theMissingIngredient Jan 16 '13 at 20:41
do you mean cylinder?? –  Santosh Linkha Jan 16 '13 at 20:46
no. I mean cone (the shape $z=\sqrt{x^2+y^2}$ ) Can't anyone help me? :( –  theMissingIngredient Jan 17 '13 at 7:36
just understood your comment. Yes,I mean the cylinder $(x-1)^2+y^2=1$ :) –  theMissingIngredient Jan 17 '13 at 7:39

Your surface is the graph of $z = f(x,y) = \sqrt{x^2+y^2}$. A standard parametrization is $$(x,y,z)=\phi(x,y)=(x,y,f(x,y))=\left(x,y,\sqrt{x^2+y^2}\right)$$ The domain $D$ in your problem is the disk $x^2 + y^2 \le 2x$ or $-\sqrt{2x-x^2}\le y \le \sqrt{2x-x^2}$, $-1\le y \le1$.

If you convert to polar coordinates, $D$ can be described by $r \le 2 \cos\theta$, $-{\pi\over2}\le \theta \le {\pi\over2}$.

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Let's write $$S_1 = \{ (x,y,z) \mid z = \sqrt{x^2 + y^2} \} \quad\text{and}\quad S_2 = \{ (x,y,z) \mid (x-1)^2 + y^2 = 1\}.$$ These are surfaces in $\mathbb R^3$, so their intersection $S$ is likely to be a curve and not a surface. Sketching shows that $S$ is an ellipse.
Since $S_2$ is a cylinder is's easy to parametrise it; global coordinates for it are $(\theta,z)$, where $x = 1 + \cos \theta$ and $y = \sin \theta$, where $0 \leq \theta \leq 2\pi$. Plugging this into the equation for $S_1$, we find that the intersection $S$ is parametrised by $$\theta \mapsto (1 + \cos\theta, \sin\theta, \sqrt{2 + 2\cos\theta}), \quad 0 \leq \theta \leq 2\pi,$$ but the only thing we're really interested in is that $z^2 = 2 + 2\cos \theta$ on $S$. This gives that the function $f$, restricted to $S$, is $$f(\theta) = 2 (2 + 2\cos\theta) - (1 + \cos\theta)^2 - \sin\theta^2 = 2 + 2\cos\theta.$$ Then your integral is $$\int_S f \, d\lambda = \int_0^{2\pi} (2 + 2\cos \theta) d\theta = 4\pi.$$