contestada

If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −2≤u≤2,−3≤v≤3, has surface area equal to 4, what is the surface area of the parametric surface given by r2(u,v)=4r1(u,v) with −2≤u≤2,−3≤v≤3?

Respuesta :

LammettHash
The area of the first surface is given to be

[tex]\displaystyle\iint_{\mathcal S_1}\left\|\frac{\partial\mathbf r_1}{\partial u}\times\frac{\partial\mathbf r_1}{\partial v}\right\|\,\mathrm du\,\mathrm dv=4[/tex]

Recall that for any scalar [tex]k[/tex] and vectors [tex]\mathbf a,\mathbf b\in\mathbb R^3[/tex],

[tex](k\mathbf a)\times\mathbf b=\mathbf a\times(k\mathbf b)=k(\mathbf a\times\mathbf b)[/tex]

which means

[tex]\left\|\dfrac{\partial\mathbf r_2}{\partial u}\times\dfrac{\partial\mathbf r_2}{\partial v}\right\|=\left\|4\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{4\partial\mathbf r_1}{\partial v}\right\|[/tex]
[tex]=16\left\|\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}\right\|[/tex]

So it follows that the area of the [tex]\mathcal S_2[/tex] is 16 times that of [tex]\mathcal S_1[/tex].

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