Mathematics: Compute the differential of surface area for the surface s described by the given parametrization. r(u, v) = e^u cos(v), e^u sin(v), uv , d = {(u, v) | 0 _ u _ 4, 0 _ v _ 2π}

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Compute the differential of surface area for the

Ответ:

jlarmee9

12.01.2022

Mathematics

With parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

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Compute the differential of surface area for the surface s described by the given parametrization.
r(u, v) = e^u cos(v), e^u sin(v), uv , d = {(u, v) | 0 < _ u < _ 4, 0 < _ v < _ 2π}

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Compute the differential of surface area for the surface s described by the given parametrization. r(u, v) = e^u cos(v), e^u sin(v), uv , d = {(u, v) | 0 < _ u < _ 4, 0 < _ v < _ 2π}

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Compute the differential of surface area for the surface s described by the given parametrization.
r(u, v) = e^u cos(v), e^u sin(v), uv , d = {(u, v) | 0 < _ u < _ 4, 0 < _ v < _ 2π}