# Calculating the mass of a Surface

Can someone help me with the following one?

I need to calculate the mass of the surface $x+y+z=a$ , $a>0$ intersected with the cylinder $x^2 + y^2 = R^2$ where the density is $u(x,y,z)= (x^2+y^2)^2 + z$ .

As far as I know, I need to calculate : $\iint_S u(x(u,v),y(u,v),z(u,v)) \,du\,dv$ where $(x(u,v), y(u,v), z(u,v))$ is the parameterization of our surface .

Can someone help me find the parameterization of such a surface?

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The intersection is not a surface, it's a curve (an ellipse, actually). So what is you density function $u$? It looks to me that if it's an area density, the curve will have mass $0$; if it's a mass attached to each point, the cylinder has infinite mass… – jathd Jan 17 '13 at 8:40
what if it is a mistake and we are actually deaing with $x^2 + y^2 \leq R^2$ ? I think this is what they meant (although they didn't write it... ). Can you help me in this particular meaning? Thanks ! – theMissingIngredient Jan 17 '13 at 8:44

Assuming the surface of interest is the intersection between the place $x+y+z=a$ and the "full" cylinder $x^2+y^2\leqslant R^2$, here's a parametrization of the surface. You can describe it as $$S = \{(x,y,z)\in\mathbf R^3 : x^2+y^2\leqslant R^2,~z=a-x-y\},$$ so you can use cylindrical coordinates: $x=r\cos(\theta)$, $y=r\sin(\theta)$, $z=a-x-y=a-r\cos(\theta)-r\sin(\theta)$ for $r\in[0,R]$ and $\theta\in[0,2\pi]$.

So the mass you're looking for is $$M = \int_{r=0}^R\int_{\theta=0}^{2\pi}(r^4+a-r\cos\theta-r\sin\theta)\,\mathrm dr\,\mathrm d\theta,$$ which shouldn't be too hard to compute.

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Thanks a lot ! !!!!!!!!!!!!!!!!!!!! – theMissingIngredient Jan 17 '13 at 10:45

Parametrizing your surface is easy: it's a plane. So you just put

$$\varphi (x,y) = (x, y, a -x-y)$$

and the limits of integration are those of the circle

$$x^2 + y^2 \leq R^2 \ .$$

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