2 years ago

The Unit Circle

Mathematics

2 answers:

Mashcka [7]2 years ago

8 0

Answer:

A. X-coordinate

Step-by-step explanation:

I calculated it logically

Pachacha [2.7K]2 years ago

8 0

Answer:

I think it's x -coordinate

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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

Bad White [126]

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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