Surface Integral - Intersection of cone and circle

Question

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?

Thanks in advance

Answer 1

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 $r \le 2 \cos\theta$, $-{\pi\over2}\le \theta \le {\pi\over2}$.

Answer 2

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 $$ \int_S f \, d\lambda = \int_0^{2\pi} (2 + 2\cos \theta) d\theta = 4\pi. $$