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I'm being asked to find the surface area of a plane defined by $x+y+z=a$ inside a cylinder defined by $x^2+y^2=a^2$ and I thought, "simple enough, I'll just use the normal vector and integrate its norm over a polar domain with $0\leq r\leq a$ and $0\leq \theta \leq 2\pi$". Sure enough, that seems to give the correct solution, according to my solution sheet:

$$\int_0^{2\pi}\int_0^a\sqrt{3}r\,dr\,d\theta = \int_0^{2\pi}\frac{\sqrt{3}}{2}a^2\,d\theta = \sqrt{3}\pi a^2$$

But let's say I want to use a parametrization $r(s,t)$ for the plane such as $r(s,t)=(a+ta+sa,-ta,-sa)$. That seems to be a correct parametrization, and it gives a normal vector $r_s \times r_t = (-a^2,-a^2,-a^2)$. Now the norm of this is $\sqrt{3}a^2$, which is gonna mess up the result obtained above. Is there a necessity for a change of variables due to the parametrization of the plane? I'm honestly out of ideas... It seems like the surface area of the plane would depend on the chosen normal vector, which is ridiculous.

(edit: $a$ is some scalar.)

user401936
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  • Shouldn't you use a unit normal for your surface integral – Triatticus Jun 24 '17 at 19:03
  • Did you write the integral? – Rafa Budría Jun 24 '17 at 19:12
  • That's for surface integrals over vector fields. If I were to make a guess, I'd say I probably can't have the norm depend on a, but I can't justify why. – user401936 Jun 24 '17 at 19:13
  • The integral w.r.t. the parametrization would simply be $\int_0^{2\pi}\int_0^a\sqrt{3}a^2r,dr,d\theta$. I'm guessing the norm of the normal vector can't depend on $a$. – user401936 Jun 24 '17 at 19:15
  • HINT:The direction cosines of normal are $ \left (\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right)$ Projected area directly increases by a factor ${\sqrt3}$, to $\pi a^2 \sqrt3, $ which can be also evaluated by usual integration of the parametrized projected ellipse. – Narasimham Jun 24 '17 at 20:23

1 Answers1

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I don't see any problem with this parametrization. Simply we have to calculate the expression for the integration surface in the new variables/parameters. From this, the relation between $s$ and $t$ is $s^2+t^2=1$. Set the correct limits for the integral and it's done,

$$S=\int_{-1}^1\int_{-\sqrt{1-t^2}}^{\sqrt{1-t^2}}\Vert r_s \times r_t\Vert dsdt =\int_{-1}^1\int_{-\sqrt{1-t^2}}^{\sqrt{1-t^2}}\sqrt{3}a^2dsdt=\sqrt{3}\pi a^2$$

Rafa Budría
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  • How would it look if I were to use a polar region? – user401936 Jun 24 '17 at 19:28
  • The region is the same, but the description in $s$ and $t$ is very diferent as in polar coordinates. The amazing thing is that all changes in a way that the calculations give the same result! – Rafa Budría Jun 24 '17 at 19:37