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RSB [31]
3 years ago
13

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

face integral (you don't need to evaluate it). ∫σ∫f(x,y,z)dS=∫R∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = 2pi/3 ∫ 0 7 ∫ 0 u^2+v^2 ∥∥∥ ∥∥∥dudv
Mathematics
1 answer:
Bad White [126]3 years ago
6 0

Looks like you have most of the details already, but you're missing one crucial piece.

\sigma is parameterized by

\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle

for 0\le u\le7 and 0\le v\le\frac{2\pi}3, and a normal vector to this surface is

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle

with norm

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}

So the integral of f(x,y,z)=x^2+y^2+z^2 is

\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}

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