Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

up vote 0 down vote accepted

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.